Use \(\sin(u\pm v)=\sin u\cos v\pm\cos u\sin v\):
\( \sin(\alpha+\beta)+\sin(\alpha-\beta)\) \(=(\sin\alpha\cos\beta+\cos\alpha\sin\beta)\) \(+(\sin\alpha\cos\beta-\cos\alpha\sin\beta)\) \(=2\sin\alpha\cos\beta. \)

Use \(\cos(u\pm v)=\cos u\cos v\mp\sin u\sin v\):
\( \cos(\alpha+\beta)+\cos(\alpha-\beta)\) \( =(\cos\alpha\cos\beta-\sin\alpha\sin\beta)\) \(+(\cos\alpha\cos\beta+\sin\alpha\sin\beta)\) \(=2\cos\alpha\cos\beta. \)

Recognize the sine–difference pattern \(\sin U\cos V-\cos U\sin V=\sin(U-V)\) with \(U=\alpha-\beta\), \(V=\alpha\):
\( \sin(\alpha-\beta)\cos\alpha-\cos(\alpha-\beta)\sin\alpha\) \(=\sin\big((\alpha-\beta)-\alpha\big)\) \(=\sin(-\beta)=-\sin\beta. \)

Start from the right-hand side and rewrite in sine/cosine:

\[ \frac{\csc\theta\,\csc\varphi}{\cot\theta+\cot\varphi} \;=\; \frac{\dfrac{1}{\sin\theta}\cdot\dfrac{1}{\sin\varphi}} {\dfrac{\cos\theta}{\sin\theta}+\dfrac{\cos\varphi}{\sin\varphi}} \]

Combine the denominator over a common denominator:

\[ =\; \frac{\dfrac{1}{\sin\theta\,\sin\varphi}} {\dfrac{\cos\theta\,\sin\varphi+\cos\varphi\,\sin\theta} {\sin\theta\,\sin\varphi}} \]

Invert and multiply:

\[ =\; \frac{1}{\cos\theta\,\sin\varphi+\cos\varphi\,\sin\theta} \]

Recognize the sine addition formula \(\sin(\theta+\varphi)=\sin\theta\cos\varphi+\cos\theta\sin\varphi\):

\[ =\; \frac{1}{\sin(\theta+\varphi)} \;=\; \csc(\theta+\varphi). \]

Hence \(\displaystyle \csc(\theta+\varphi)\equiv \dfrac{\csc\theta\,\csc\varphi}{\cot\theta+\cot\varphi}\).

Expand each cosine:

\[ \cos(5x+x) = \cos 5x\cos x - \sin 5x\sin x, \] \[ \cos(5x-x) = \cos 5x\cos x + \sin 5x\sin x. \]

Add the two results:

\[ \cos 6x + \cos 4x = (\cos 5x\cos x - \sin 5x\sin x) + (\cos 5x\cos x + \sin 5x\sin x) = 2\cos 5x\cos x. \]

Hence proved.

Start from the identity \(\cos(x-y)=\cos x\cos y+\sin x\sin y\).

Compute \(p^2+q^2\):

\[ p^2+q^2 =(\sin x+\sin y)^2 + (\cos x+\cos y)^2 \] \[ =(\sin^2x+\sin^2y + 2\sin x\sin y) +(\cos^2x+\cos^2y + 2\cos x\cos y) \] \[ =( \sin^2x+\cos^2x ) + ( \sin^2y+\cos^2y ) + 2(\sin x\sin y+\cos x\cos y) \] \[ =1+1+2\cos(x-y) =2+2\cos(x-y). \]

Solve for \(\cos(x-y)\):

\[ \cos(x-y)=\frac{p^2+q^2-2}{2}. \]

Result: \(\displaystyle \boxed{\cos(x-y)=\tfrac{p^2+q^2-2}{2}}\).

(a) Use \(\cos(A+B)=\cos A\cos B-\sin A\sin B\):

\[ \cos 3\theta=\cos(2\theta+\theta) =(\cos2\theta)\cos\theta-(\sin2\theta)\sin\theta . \] With \(\cos2\theta=2\cos^2\theta-1\) and \(\sin2\theta=2\sin\theta\cos\theta\), \[ \cos 3\theta =(2\cos^2\theta-1)\cos\theta-2\sin\theta\cos\theta\sin\theta =4\cos^3\theta-3\cos\theta. \]

(b) Substitute the identity from (a) and \(\cos2\theta=2\cos^2\theta-1\):

\[ (4\cos^3\theta-3\cos\theta)+\cos\theta(2\cos^2\theta-1) =\cos^2\theta . \] \[ \Rightarrow\; 6\cos^3\theta-4\cos\theta=\cos^2\theta \;\Rightarrow\; 6\cos^3\theta-\cos^2\theta-4\cos\theta=0 \] \[ \Rightarrow\; \cos\theta\,(6\cos^2\theta-\cos\theta-4)=0 . \]

Either \(\cos\theta=0\Rightarrow \theta=90^\circ\), or \(6\cos^2\theta-\cos\theta-4=0\). Solving the quadratic in \(c=\cos\theta\): \[ c=\frac{1\pm\sqrt{1+96}}{12} =\frac{1\pm\sqrt{97}}{12} \approx 0.9041\;\text{ or }\;-0.7374. \] Hence \[ \theta=\cos^{-1}(0.9041)\approx 25.3^\circ,\qquad \theta=\cos^{-1}(-0.7374)\approx 137.5^\circ, \] together with \(\theta=90^\circ\) from \(\cos\theta=0\).

Expand each term using \(\sin(u\pm v)=\sin u\cos v\pm\cos u\sin v\) and \(\cos(u\pm v)=\cos u\cos v\mp\sin u\sin v\):

\[ \sin\!\left(x+\tfrac{\pi}{6}\right) -\sin\!\left(x-\tfrac{\pi}{6}\right) =(\sin x\cos\tfrac{\pi}{6}+\cos x\sin\tfrac{\pi}{6}) -(\sin x\cos\tfrac{\pi}{6}-\cos x\sin\tfrac{\pi}{6}) \] \[ =2\cos x\sin\tfrac{\pi}{6} =2\cos x\cdot\tfrac12 =\cos x. \]

\[ \cos\!\left(x+\tfrac{\pi}{3}\right) -\cos\!\left(x-\tfrac{\pi}{3}\right) =(\cos x\cos\tfrac{\pi}{3}-\sin x\sin\tfrac{\pi}{3}) -(\cos x\cos\tfrac{\pi}{3}+\sin x\sin\tfrac{\pi}{3}) \] \[ =-2\sin x\sin\tfrac{\pi}{3} =-2\sin x\cdot\tfrac{\sqrt3}{2} =-\sqrt3\,\sin x. \]

Equate LHS and RHS: \[ \cos x=-\sqrt3\,\sin x \quad\Rightarrow\quad \tan x=\frac{\sin x}{\cos x}=-\frac{1}{\sqrt3}. \]

(b) Since \(\tan x = -\tfrac{1}{\sqrt3}\) (reference angle \(=\tfrac{\pi}{6}\)) and tangent is negative in quadrants II and IV, \[ x = \pi-\tfrac{\pi}{6}=\tfrac{5\pi}{6}, \qquad x = 2\pi-\tfrac{\pi}{6}=\tfrac{11\pi}{6}. \]

Final answers

(a) \(\displaystyle \tan x=-\frac{1}{\sqrt3}\).
(b) \(\displaystyle x=\frac{5\pi}{6},\;\frac{11\pi}{6}\) on \(0\le x\le 2\pi\).

Use the tangent addition formula: \( \tan(x+45^\circ)=\frac{\tan x+1}{\,1-\tan x\,}. \) Also \(\cot x=\dfrac{1}{\tan x}\).

Let \(t=\tan x\). Then \[ \frac{t+1}{1-t}=2\cdot\frac{1}{t} \;\Rightarrow\; t(t+1)=2(1-t) \] \[ \Rightarrow\; t^2+t=2-2t \;\Rightarrow\; t^2+3t-2=0. \]

Solve the quadratic: \[ t=\frac{-3\pm\sqrt{9+8}}{2} =\frac{-3\pm\sqrt{17}}{2} \approx 0.5616\;\text{ or }\;-3.5616. \]

\[ \tan x\approx0.5616 \Rightarrow x\approx \arctan(0.5616)=29.3^\circ, \] \[ \tan x\approx-3.5616 \Rightarrow x\text{ in QII } =180^\circ-\arctan(3.5616)=105.7^\circ. \]

Final answer

\(\boxed{x\approx29.3^\circ,\;105.7^\circ}\)

Expand the left side with \(\cos(u-v)=\cos u\cos v+\sin u\sin v\):

\[ \cos(\theta-60^\circ) =\cos\theta\cos60^\circ+\sin\theta\sin60^\circ =\tfrac12\cos\theta+\tfrac{\sqrt3}{2}\sin\theta. \]

Set equal to \(3\sin\theta\) and rearrange:

\[ \tfrac12\cos\theta+\tfrac{\sqrt3}{2}\sin\theta=3\sin\theta \;\Rightarrow\; \tfrac12\cos\theta=\Big(3-\tfrac{\sqrt3}{2}\Big)\sin\theta. \] \[ \cos\theta=(6-\sqrt3)\,\sin\theta \;\Rightarrow\; \tan\theta=\frac{\sin\theta}{\cos\theta} =\frac{1}{\,6-\sqrt3\,}. \]

Evaluate the principal angle:

\[ \theta=\arctan\!\Big(\frac{1}{6-\sqrt3}\Big) \approx \arctan(0.2343)=13.2^\circ. \]

Tangent is positive in quadrants I and III, so add \(180^\circ\):

\[ \theta=13.2^\circ,\quad 13.2^\circ+180^\circ=193.2^\circ. \]

1. Use the identity \(\cot 2\theta = \frac{1}{\tan 2\theta}\) and \(\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}\) to express \(\cot 2\theta\) in terms of \(\tan \theta\):

\(\cot 2\theta = \frac{1 - \tan^2 \theta}{2 \tan \theta}\)

2. Use the identity \(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\) to express \(\tan(\theta + 45^\circ)\):

\(\tan(\theta + 45^\circ) = \frac{\tan \theta + 1}{1 - \tan \theta}\)

3. Substitute these into the original equation:

\(\tan \theta \cdot \frac{\tan \theta + 1}{1 - \tan \theta} = 2 \cdot \frac{1 - \tan^2 \theta}{2 \tan \theta}\)

4. Simplify and rearrange to form a quadratic equation:

\(2 \tan^2 \theta + \tan \theta - 1 = 0\)

5. Solve the quadratic equation using the quadratic formula \(\tan \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2, b = 1, c = -1\):

\(\tan \theta = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}\)

6. This gives \(\tan \theta = \frac{1}{2}\) or \(\tan \theta = -1\).

7. Since \(0^\circ < \theta < 90^\circ\), \(\tan \theta = \frac{1}{2}\) is valid.

8. Therefore, \(\theta = \tan^{-1}\left(\frac{1}{2}\right) \approx 26.6^\circ\).

1. Use the tangent addition and subtraction formulas:

\(\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}\)

2. For \(\tan(\theta + 60^\circ)\):

\(\tan(\theta + 60^\circ) = \frac{\tan \theta + \sqrt{3}}{1 - \tan \theta \cdot \sqrt{3}}\)

3. For \(\tan(60^\circ - \theta)\):

\(\tan(60^\circ - \theta) = \frac{\sqrt{3} - \tan \theta}{1 + \tan \theta \cdot \sqrt{3}}\)

4. Set the equation:

\(\frac{\tan \theta + \sqrt{3}}{1 - \tan \theta \cdot \sqrt{3}} = 2 + \frac{\sqrt{3} - \tan \theta}{1 + \tan \theta \cdot \sqrt{3}}\)

5. Clear the fractions by multiplying through by \((1 - \tan \theta \cdot \sqrt{3})(1 + \tan \theta \cdot \sqrt{3})\).

6. Simplify and rearrange to form a quadratic equation:

\(3\tan^2 \theta + 4\tan \theta - 1 = 0\)

7. Solve the quadratic equation using the quadratic formula:

\(\tan \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

where \(a = 3, b = 4, c = -1\).

8. Calculate:

\(\tan \theta = \frac{-4 \pm \sqrt{16 + 12}}{6} = \frac{-4 \pm \sqrt{28}}{6} = \frac{-4 \pm 2\sqrt{7}}{6}\)

9. Simplify:

\(\tan \theta = \frac{-2 \pm \sqrt{7}}{3}\)

10. Find \(\theta\) using \(\tan^{-1}\):

\(\theta \approx 12.1^\circ\) and \(\theta \approx 122.9^\circ\).

(a) Start by expressing the left-hand side as a single fraction:

\(\frac{\cos 3x}{\sin x} + \frac{\sin 3x}{\cos x} = \frac{\cos 3x \cdot \cos x + \sin 3x \cdot \sin x}{\sin x \cdot \cos x}\)

Use the angle addition formula \(\cos(A - B) = \cos A \cos B + \sin A \sin B\):

\(= \frac{\cos(3x - x)}{\sin x \cos x} = \frac{\cos 2x}{\sin x \cos x}\)

Use the double angle identity \(\sin 2x = 2 \sin x \cos x\):

\(= \frac{\cos 2x}{\frac{1}{2} \sin 2x} = \frac{2 \cos 2x}{\sin 2x} = 2 \cot 2x\)

Thus, the identity is proven.

(b) Set \(\frac{\cos 3x}{\sin x} + \frac{\sin 3x}{\cos x} = 4\):

\(2 \cot 2x = 4\) \(\cot 2x = 2\)

Then, \(\tan 2x = \frac{1}{2}\).

Find \(2x\):

\(2x = \tan^{-1}\left(\frac{1}{2}\right) \approx 0.464 \text{ or } 0.464 + n\pi\)

For \(0 < x < \pi\), solve for \(x\):

\(x = \frac{0.464}{2} \approx 0.232\)

And for the next solution:

\(x = \frac{0.464 + \pi}{2} \approx 1.80\)

Thus, \(x \approx 0.232, 1.80\).

(i) Start by expanding \(\tan(2x + x)\) using the angle addition formula:

\(\tan(2x + x) = \frac{\tan 2x + \tan x}{1 - \tan 2x \tan x}\)

Using \(\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}\), substitute into the equation:

\(\tan 3x = \frac{\frac{2 \tan x}{1 - \tan^2 x} + \tan x}{1 - \frac{2 \tan x}{1 - \tan^2 x} \tan x}\)

Simplify the expression:

\(\tan 3x = \frac{2 \tan x + \tan x (1 - \tan^2 x)}{1 - 2 \tan^2 x}\)

\(\tan 3x = \frac{3 \tan x - \tan^3 x}{1 - 2 \tan^2 x}\)

Given \(\tan 3x = 3 \cot x = \frac{3}{\tan x}\), equate and simplify:

\(\frac{3 \tan x - \tan^3 x}{1 - 2 \tan^2 x} = \frac{3}{\tan x}\)

Cross-multiply and simplify:

\(3 \tan^2 x - \tan^4 x = 3 - 6 \tan^2 x\)

Rearrange to obtain:

\(\tan^4 x - 12 \tan^2 x + 3 = 0\)

(ii) Solve \(\tan^4 x - 12 \tan^2 x + 3 = 0\) by substituting \(y = \tan^2 x\):

\(y^2 - 12y + 3 = 0\)

Use the quadratic formula:

\(y = \frac{12 \pm \sqrt{144 - 12}}{2}\)

\(y = \frac{12 \pm \sqrt{132}}{2}\)

\(y = 6 \pm \sqrt{33}\)

Since \(y = \tan^2 x\), solve for \(x\):

\(\tan x = \sqrt{6 + \sqrt{33}}\) or \(\tan x = \sqrt{6 - \sqrt{33}}\)

Calculate \(x\) for \(0^\circ < x < 90^\circ\):

\(x = 26.8^\circ\) and \(x = 73.7^\circ\)

1. Start with the equation:

\(\cot \theta - \cot(\theta + 45^\circ) = 3\)

2. Use the identity \(\cot(\theta + 45^\circ) = \frac{1 - \tan \theta}{\tan \theta + 1}\).

3. Substitute \(\cot \theta = \frac{1}{\tan \theta}\) and \(\cot(\theta + 45^\circ) = \frac{1 - \tan \theta}{\tan \theta + 1}\) into the equation:

\(\frac{1}{\tan \theta} - \frac{1 - \tan \theta}{\tan \theta + 1} = 3\)

4. Simplify and rearrange to form a quadratic equation:

\(1 + \tan \theta - \tan \theta(1 - \tan \theta) = 3 \tan \theta(1 + \tan \theta)\)

5. Expand and simplify:

\(2\tan^2 \theta + 3\tan \theta - 1 = 0\)

6. Solve the quadratic equation using the quadratic formula:

\(\tan \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

where \(a = 2\), \(b = 3\), \(c = -1\).

7. Calculate:

\(\tan \theta = \frac{-3 \pm \sqrt{3^2 - 4 \times 2 \times (-1)}}{2 \times 2}\)

\(\tan \theta = \frac{-3 \pm \sqrt{9 + 8}}{4}\)

\(\tan \theta = \frac{-3 \pm \sqrt{17}}{4}\)

8. Find \(\theta\) for \(0^\circ < \theta < 180^\circ\):

\(\theta = 15.7^\circ\) or \(\theta = 119.3^\circ\).

(i) Start with the given equation:

\(\sin(\theta + 45^\circ) + 2 \cos(\theta + 60^\circ) = 3 \cos \theta\)

Use angle addition formulas:

\(\sin(\theta + 45^\circ) = \sin \theta \cos 45^\circ + \cos \theta \sin 45^\circ\)

\(\cos(\theta + 60^\circ) = \cos \theta \cos 60^\circ - \sin \theta \sin 60^\circ\)

Substitute \(\cos 45^\circ = \sin 45^\circ = \frac{\sqrt{2}}{2}\), \(\cos 60^\circ = \frac{1}{2}\), and \(\sin 60^\circ = \frac{\sqrt{3}}{2}\):

\(\frac{\sqrt{2}}{2} \sin \theta + \frac{\sqrt{2}}{2} \cos \theta + 2 \left( \frac{1}{2} \cos \theta - \frac{\sqrt{3}}{2} \sin \theta \right) = 3 \cos \theta\)

Simplify:

\(\frac{\sqrt{2}}{2} \sin \theta + \frac{\sqrt{2}}{2} \cos \theta + \cos \theta - \sqrt{3} \sin \theta = 3 \cos \theta\)

Combine terms:

\(\left( \frac{\sqrt{2}}{2} - \sqrt{3} \right) \sin \theta + \left( \frac{\sqrt{2}}{2} + 1 \right) \cos \theta = 3 \cos \theta\)

Rearrange:

\(\left( \frac{\sqrt{2}}{2} - \sqrt{3} \right) \sin \theta = \left( 3 - \frac{\sqrt{2}}{2} - 1 \right) \cos \theta\)

\(\left( \frac{\sqrt{2}}{2} - \sqrt{3} \right) \sin \theta = \left( 2 - \frac{\sqrt{2}}{2} \right) \cos \theta\)

Divide both sides by \(\cos \theta\):

\(\tan \theta = \frac{2\sqrt{2} - 1}{1 - \sqrt{6}}\)

(ii) Solve for \(\theta\):

Using the expression for \(\tan \theta\), find \(\theta\) in the range \(0^\circ < \theta < 360^\circ\).

\(\theta = 128.4^\circ\) and \(\theta = 308.4^\circ\)

To solve the equation \(\sin(\theta - 30^\circ) + \cos \theta = 2 \sin \theta\), we start by using the angle subtraction identity:

\(\sin(\theta - 30^\circ) = \sin \theta \cos 30^\circ - \cos \theta \sin 30^\circ\)

Substitute the values \(\cos 30^\circ = \frac{\sqrt{3}}{2}\) and \(\sin 30^\circ = \frac{1}{2}\):

\(\sin \theta \cdot \frac{\sqrt{3}}{2} - \cos \theta \cdot \frac{1}{2} + \cos \theta = 2 \sin \theta\)

Simplify the equation:

\(\frac{\sqrt{3}}{2} \sin \theta - \frac{1}{2} \cos \theta + \cos \theta = 2 \sin \theta\)

Combine like terms:

\(\frac{\sqrt{3}}{2} \sin \theta + \frac{1}{2} \cos \theta = 2 \sin \theta\)

Rearrange to form:

\(\frac{\sqrt{3}}{2} \sin \theta - 2 \sin \theta = -\frac{1}{2} \cos \theta\)

Factor out \(\sin \theta\):

\(\sin \theta \left( \frac{\sqrt{3}}{2} - 2 \right) = -\frac{1}{2} \cos \theta\)

Divide both sides by \(\cos \theta\):

\(\tan \theta = \frac{1}{4 - \sqrt{3}}\)

Calculate \(\theta\):

\(\theta = \tan^{-1}\left(\frac{1}{4 - \sqrt{3}}\right) \approx 23.8^\circ\)

Thus, the solution is \(\theta = 23.8^\circ\).

We solve the equation \(\cot \theta + \cot(\theta + 45^\circ) = 2\) using the addition formula for tangent.

Let \(t = \tan \theta\). Using the identity

\[ \tan(\theta + 45^\circ) = \frac{\tan \theta + \tan 45^\circ}{1 - \tan \theta \tan 45^\circ} = \frac{t + 1}{1 - t}, \] so \[ \cot(\theta + 45^\circ) = \frac{1}{\tan(\theta + 45^\circ)} = \frac{1 - t}{t + 1}. \]

Also, \(\cot \theta = \frac{1}{t}\). Substituting these into the original equation gives

\[ \frac{1}{t} + \frac{1 - t}{t + 1} = 2. \]

Multiply through by \(t(t+1)\):

\[ (t+1) + t(1 - t) = 2t(t+1). \]

Expanding and simplifying:

\[ t + 1 + t - t^2 = 2t^2 + 2t \quad \Rightarrow \quad 2t - t^2 + 1 = 2t^2 + 2t. \]

Bring all terms to one side:

\[ 0 = 3t^2 - 1 \quad \Rightarrow \quad t^2 = \frac{1}{3}. \]

So

\[ t = \pm \frac{1}{\sqrt{3}}. \]

This gives \(\tan \theta = \frac{1}{\sqrt{3}}\) or \(\tan \theta = -\frac{1}{\sqrt{3}}\).

In degrees:

• \(\tan \theta = \frac{1}{\sqrt{3}}\) gives \(\theta = 30^\circ + 180^\circ k\).
• \(\tan \theta = -\frac{1}{\sqrt{3}}\) gives \(\theta = 150^\circ + 180^\circ k\).

For \(0^\circ < \theta < 180^\circ\), the solutions are \(\boxed{\theta = 30^\circ \text{ and } \theta = 150^\circ}\).

Part (i):

Given the equation:

\(\sin(x - 60^\circ) = 3 \cos(x - 45^\circ)\)

Using the angle subtraction identities:

\(\sin(x - 60^\circ) = \sin x \cos 60^\circ - \cos x \sin 60^\circ\)

\(\cos(x - 45^\circ) = \cos x \cos 45^\circ + \sin x \sin 45^\circ\)

Substitute the exact values:

\(\sin x \cdot \frac{1}{2} - \cos x \cdot \frac{\sqrt{3}}{2} = 3(\cos x \cdot \frac{\sqrt{2}}{2} + \sin x \cdot \frac{\sqrt{2}}{2})\)

Rearrange and simplify:

\(\frac{1}{2} \sin x - \frac{\sqrt{3}}{2} \cos x = \frac{3\sqrt{2}}{2} \cos x + \frac{3\sqrt{2}}{2} \sin x\)

Combine terms:

\(\left(\frac{1}{2} - \frac{3\sqrt{2}}{2}\right) \sin x = \left(\frac{3\sqrt{2}}{2} + \frac{\sqrt{3}}{2}\right) \cos x\)

\(\sin x \left(1 - 3\sqrt{2}\right) = \cos x \left(3\sqrt{2} + \sqrt{3}\right)\)

Divide both sides by \(\cos x\):

\(\tan x = \frac{-(6+\sqrt{6})}{(6-\sqrt{2})}\)

Part (ii):

Using the result from part (i), solve for \(x\):

\(x = 118.5^\circ, 298.5^\circ\)

(i) To prove the identity \(\tan(45^\circ + x) + \tan(45^\circ - x) \equiv 2 \sec 2x\):

Use the tangent addition and subtraction formulas:

\(\tan(45^\circ + x) = \frac{\tan 45^\circ + \tan x}{1 - \tan 45^\circ \tan x} = \frac{1 + \tan x}{1 - \tan x}\)

\(\tan(45^\circ - x) = \frac{\tan 45^\circ - \tan x}{1 + \tan 45^\circ \tan x} = \frac{1 - \tan x}{1 + \tan x}\)

Add the two expressions:

\(\tan(45^\circ + x) + \tan(45^\circ - x) = \frac{1 + \tan x}{1 - \tan x} + \frac{1 - \tan x}{1 + \tan x}\)

Combine into a single fraction:

\(= \frac{(1 + \tan x)^2 + (1 - \tan x)^2}{(1 - \tan x)(1 + \tan x)}\)

Simplify the numerator:

\(= \frac{1 + 2\tan^2 x + 1}{1 - \tan^2 x} = \frac{2(1 + \tan^2 x)}{1 - \tan^2 x}\)

Using the identity \(1 + \tan^2 x = \sec^2 x\), we have:

\(= \frac{2 \sec^2 x}{1 - \tan^2 x} = 2 \sec 2x\)

Thus, the identity is proven.

(ii) To sketch the graph of \(y = \tan(45^\circ + x) + \tan(45^\circ - x)\) for \(0^\circ \leq x \leq 90^\circ\):

The function has an asymptote at \(x = 45^\circ\) because \(\tan(45^\circ + x)\) and \(\tan(45^\circ - x)\) become undefined.

The graph consists of two branches approaching the asymptote at \(x = 45^\circ\).

1. Use the angle addition and subtraction formulas:

\(\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}\)

2. Apply the formulas:

\(\tan(\theta + 45^\circ) = \frac{\tan \theta + 1}{1 - \tan \theta}\)

\(\tan(\theta - 45^\circ) = \frac{\tan \theta - 1}{1 + \tan \theta}\)

3. Substitute into the equation:

\(\frac{\tan \theta + 1}{1 - \tan \theta} - 2 \left(\frac{\tan \theta - 1}{1 + \tan \theta}\right) = 4\)

4. Simplify and solve for \(\tan \theta\):

Multiply through by \((1 - \tan \theta)(1 + \tan \theta)\):

\((\tan \theta + 1)(1 + \tan \theta) - 2(\tan \theta - 1)(1 - \tan \theta) = 4(1 - \tan^2 \theta)\)

5. Simplify further:

\(\tan^2 \theta + 2\tan \theta + 1 - 2(\tan^2 \theta - 1) = 4 - 4\tan^2 \theta\)

\(\tan^2 \theta + 2\tan \theta + 1 - 2\tan^2 \theta + 2 = 4 - 4\tan^2 \theta\)

6. Rearrange to form a quadratic equation:

\(7\tan^2 \theta - 2\tan \theta - 1 = 0\)

7. Solve the quadratic equation using the quadratic formula:

\(\tan \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

where \(a = 7, b = -2, c = -1\).

8. Calculate:

\(\tan \theta = \frac{2 \pm \sqrt{(-2)^2 - 4 \times 7 \times (-1)}}{2 \times 7}\)

\(\tan \theta = \frac{2 \pm \sqrt{4 + 28}}{14}\)

\(\tan \theta = \frac{2 \pm \sqrt{32}}{14}\)

\(\tan \theta = \frac{2 \pm 4\sqrt{2}}{14}\)

\(\tan \theta = \frac{1 \pm 2\sqrt{2}}{7}\)

9. Find \(\theta\) using \(\tan^{-1}\):

\(\theta = 28.7^\circ\) and \(\theta = 165.4^\circ\).

1. Use the identity for \(\sin(A + 45^\circ)\):

\(\sin A \cos 45^\circ + \cos A \sin 45^\circ = 2\sqrt{2} \cos A\)

2. Simplify using \(\cos 45^\circ = \sin 45^\circ = \frac{\sqrt{2}}{2}\):

\(\sin A \cdot \frac{\sqrt{2}}{2} + \cos A \cdot \frac{\sqrt{2}}{2} = 2\sqrt{2} \cos A\)

3. Divide through by \(\cos A\):

\(\tan A \cdot \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = 2\sqrt{2}\)

4. Solve for \(\tan A\):

\(\tan A = 3\)

5. Use the identity \(\sec^2 B = 1 + \tan^2 B\) in the second equation:

\(4(1 + \tan^2 B) + 5 = 12 \tan B\)

6. Simplify and solve the quadratic equation:

\(4\tan^2 B - 12\tan B + 9 = 0\)

7. Factor or use the quadratic formula to find:

\(\tan B = \frac{3}{2}\)

8. Use the identity for \(\tan(A - B)\):

\(\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}\)

9. Substitute \(\tan A = 3\) and \(\tan B = \frac{3}{2}\):

\(\tan(A - B) = \frac{3 - \frac{3}{2}}{1 + 3 \cdot \frac{3}{2}}\)

10. Simplify:

\(\tan(A - B) = \frac{\frac{3}{2}}{1 + \frac{9}{2}} = \frac{\frac{3}{2}}{\frac{11}{2}} = \frac{3}{11}\)

Given:

\(\tan(\theta - \phi) = 3\)

\(\tan \theta + \tan \phi = 1\)

Using the tangent subtraction formula:

\(\tan(\theta - \phi) = \frac{\tan \theta - \tan \phi}{1 + \tan \theta \tan \phi} = 3\)

Substitute \(\tan \theta + \tan \phi = 1\) into the equation:

\(\frac{\tan \theta - (1 - \tan \theta)}{1 + \tan \theta (1 - \tan \theta)} = 3\)

Simplify:

\(\frac{2\tan \theta - 1}{1 + \tan \theta - \tan^2 \theta} = 3\)

Cross-multiply and simplify:

\(2\tan \theta - 1 = 3(1 + \tan \theta - \tan^2 \theta)\)

\(2\tan \theta - 1 = 3 + 3\tan \theta - 3\tan^2 \theta\)

Rearrange to form a quadratic equation:

\(3\tan^2 \theta - \tan \theta - 4 = 0\)

Solving this quadratic equation gives:

\(\tan \theta = 1.5 \quad \text{or} \quad \tan \theta = -\frac{4}{3}\)

For \(\tan \theta = 1.5\), \(\theta \approx 56.3^\circ\) or \(236.3^\circ\) (only \(56.3^\circ\) is valid).

For \(\tan \theta = -\frac{4}{3}\), \(\theta \approx 126.9^\circ\) or \(306.9^\circ\) (only \(126.9^\circ\) is valid).

Using \(\tan \theta + \tan \phi = 1\), find \(\phi\):

For \(\theta = 56.3^\circ\), \(\phi \approx 161.6^\circ\).

For \(\theta = 126.9^\circ\), \(\phi \approx 63.4^\circ\).

Thus, the possible pairs are \((\theta, \phi) = (135^\circ, 63.4^\circ)\) and \((53.1^\circ, 161.6^\circ)\).

(i) Use the sum-to-product identities:

\(\cos(\theta - 60^\circ) + \cos(\theta + 60^\circ) = 2 \cos \left( \frac{2\theta}{2} \right) \cos \left( \frac{-120^\circ}{2} \right)\)

\(= 2 \cos \theta \cos(-60^\circ)\)

Since \(\cos(-60^\circ) = \cos(60^\circ) = \frac{1}{2}\), we have:

\(2 \cos \theta \cdot \frac{1}{2} = \cos \theta\)

Thus, \(\cos(\theta - 60^\circ) + \cos(\theta + 60^\circ) = \cos \theta\).

(ii) Start with:

\(\frac{\cos(2x - 60^\circ) + \cos(2x + 60^\circ)}{\cos(x - 60^\circ) + \cos(x + 60^\circ)} = 3\)

Using the identity \(\cos A + \cos B = 2 \cos \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)\), we have:

\(\cos(2x - 60^\circ) + \cos(2x + 60^\circ) = 2 \cos(2x) \cos(60^\circ) = \cos(2x)\)

\(\cos(x - 60^\circ) + \cos(x + 60^\circ) = 2 \cos(x) \cos(60^\circ) = \cos(x)\)

Thus, \(\frac{\cos(2x)}{\cos(x)} = 3\), implying \(\cos(2x) = 3 \cos(x)\).

Using the double angle identity \(\cos(2x) = 2 \cos^2(x) - 1\), we get:

\(2 \cos^2(x) - 1 = 3 \cos(x)\)

\(2 \cos^2(x) - 3 \cos(x) - 1 = 0\)

Solving this quadratic equation for \(\cos(x)\):

Let \(y = \cos(x)\), then:

\(2y^2 - 3y - 1 = 0\)

Using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = -3\), \(c = -1\):

\(y = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot (-1)}}{4}\)

\(y = \frac{3 \pm \sqrt{9 + 8}}{4}\)

\(y = \frac{3 \pm \sqrt{17}}{4}\)

Since \(\cos(x)\) must be between -1 and 1, the valid solution is:

\(\cos(x) = \frac{1}{4}(3 - \sqrt{17})\)

(i) Expand \(\sin(2\theta + \theta)\) using the angle addition formula:

\(\sin(2\theta + \theta) = \sin 2\theta \cos \theta + \cos 2\theta \sin \theta\)

Using double angle identities:

\(\sin 2\theta = 2 \sin \theta \cos \theta\)

\(\cos 2\theta = 1 - 2 \sin^2 \theta\)

Substitute these into the expansion:

\(\sin 3\theta = (2 \sin \theta \cos \theta) \cos \theta + (1 - 2 \sin^2 \theta) \sin \theta\)

\(= 2 \sin \theta \cos^2 \theta + \sin \theta - 2 \sin^3 \theta\)

\(= \sin \theta (2 \cos^2 \theta + 1 - 2 \sin^2 \theta)\)

\(= \sin \theta (2(1 - \sin^2 \theta) + 1 - 2 \sin^2 \theta)\)

\(= \sin \theta (3 - 4 \sin^2 \theta)\)

\(= 3 \sin \theta - 4 \sin^3 \theta\)

This proves the identity.

(ii) Substitute \(x = \frac{2 \sin \theta}{\sqrt{3}}\) into the equation:

\(x^3 - x + \frac{1}{6}\sqrt{3} = 0\)

\(\left(\frac{2 \sin \theta}{\sqrt{3}}\right)^3 - \frac{2 \sin \theta}{\sqrt{3}} + \frac{1}{6}\sqrt{3} = 0\)

\(\frac{8 \sin^3 \theta}{3\sqrt{3}} - \frac{2 \sin \theta}{\sqrt{3}} + \frac{1}{6}\sqrt{3} = 0\)

Multiply through by \(3\sqrt{3}\):

\(8 \sin^3 \theta - 6 \sin \theta + \frac{1}{2} \cdot 3 = 0\)

\(8 \sin^3 \theta - 6 \sin \theta + \frac{3}{2} = 0\)

\(8 \sin^3 \theta - 6 \sin \theta = -\frac{3}{2}\)

\(\sin 3\theta = 3 \sin \theta - 4 \sin^3 \theta = \frac{3}{4}\)

This confirms the form \(\sin 3\theta = \frac{3}{4}\).

(iii) Solve \(x^3 - x + \frac{1}{6}\sqrt{3} = 0\):

Using numerical methods or a calculator, find:

\(x = 0.322, 0.799, -1.12\)

Part (i):

Start with the given equation:

\(\tan(x - 60^\circ) + \cot x = \sqrt{3}\)

Using the identity \(\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}\), we have:

\(\tan(x - 60^\circ) = \frac{\tan x - \sqrt{3}}{1 + \sqrt{3} \tan x}\)

Substitute into the equation:

\(\frac{\tan x - \sqrt{3}}{1 + \sqrt{3} \tan x} + \frac{1}{\tan x} = \sqrt{3}\)

Multiply through by \(\tan x (1 + \sqrt{3} \tan x)\):

\((\tan x - \sqrt{3}) + (1 + \sqrt{3} \tan x) = \sqrt{3} \tan x (1 + \sqrt{3} \tan x)\)

Simplify and rearrange:

\(\tan x - \sqrt{3} + 1 + \sqrt{3} \tan x = \sqrt{3} \tan x + 3 \tan^2 x\)

\(2 \tan x - \sqrt{3} + 1 = 3 \tan^2 x\)

Rearrange to form:

\(2 \tan^2 x + \sqrt{3} \tan x - 1 = 0\)

Part (ii):

Solve the quadratic equation:

\(2 \tan^2 x + \sqrt{3} \tan x - 1 = 0\)

Let \(u = \tan x\), then:

\(2u^2 + \sqrt{3}u - 1 = 0\)

Using the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = \sqrt{3}\), \(c = -1\):

\(u = \frac{-\sqrt{3} \pm \sqrt{(\sqrt{3})^2 - 4 \times 2 \times (-1)}}{2 \times 2}\)

\(u = \frac{-\sqrt{3} \pm \sqrt{3 + 8}}{4}\)

\(u = \frac{-\sqrt{3} \pm \sqrt{11}}{4}\)

Calculate \(\tan x\) and find \(x\):

\(x = 21.6^\circ\) and \(x = 128.4^\circ\)

To solve the equation \(\cos(x + 30^\circ) = 2 \cos x\), we use the cosine addition formula:

\(\cos(x + 30^\circ) = \cos x \cos 30^\circ - \sin x \sin 30^\circ.\)

Substitute \(\cos 30^\circ = \frac{\sqrt{3}}{2}\) and \(\sin 30^\circ = \frac{1}{2}\):

\(\frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x = 2 \cos x.\)

Rearrange to form:

\(\frac{\sqrt{3}}{2} \cos x - 2 \cos x = \frac{1}{2} \sin x.\)

Simplify:

\((\frac{\sqrt{3}}{2} - 2) \cos x = \frac{1}{2} \sin x.\)

Divide both sides by \(\cos x\):

\(\tan x = \frac{\frac{1}{2}}{2 - \frac{\sqrt{3}}{2}}.\)

Simplify the expression:

\(\tan x = \sqrt{3} - 4.\)

Find \(x\) using the arctangent function:

\(x = \tan^{-1}(\sqrt{3} - 4).\)

Calculate the solutions within the interval \(-180^\circ < x < 180^\circ\):

\(x \approx -66.2^\circ \text{ and } x \approx 113.8^\circ.\)

To solve \(\sin(\theta + 45^\circ) = 2 \cos(\theta - 30^\circ)\), we use the angle addition formulas:

\(\sin(\theta + 45^\circ) = \sin \theta \cos 45^\circ + \cos \theta \sin 45^\circ\)

\(\cos(\theta - 30^\circ) = \cos \theta \cos 30^\circ + \sin \theta \sin 30^\circ\)

Substitute these into the equation:

\(\sin \theta \cos 45^\circ + \cos \theta \sin 45^\circ = 2(\cos \theta \cos 30^\circ + \sin \theta \sin 30^\circ)\)

Using \(\cos 45^\circ = \sin 45^\circ = \frac{\sqrt{2}}{2}\), \(\cos 30^\circ = \frac{\sqrt{3}}{2}\), and \(\sin 30^\circ = \frac{1}{2}\), we have:

\(\frac{\sqrt{2}}{2} \sin \theta + \frac{\sqrt{2}}{2} \cos \theta = 2 \left( \frac{\sqrt{3}}{2} \cos \theta + \frac{1}{2} \sin \theta \right)\)

Simplify:

\(\frac{\sqrt{2}}{2} \sin \theta + \frac{\sqrt{2}}{2} \cos \theta = \sqrt{3} \cos \theta + \sin \theta\)

Rearrange terms:

\(\frac{\sqrt{2}}{2} \sin \theta - \sin \theta = \sqrt{3} \cos \theta - \frac{\sqrt{2}}{2} \cos \theta\)

\(\left( \frac{\sqrt{2}}{2} - 1 \right) \sin \theta = \left( \sqrt{3} - \frac{\sqrt{2}}{2} \right) \cos \theta\)

Divide both sides by \(\cos \theta\):

\(\tan \theta = \frac{\sqrt{3} - \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2} - 1}\)

Calculate \(\theta\):

\(\theta = \tan^{-1}\left( \frac{\sqrt{3} - \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2} - 1} \right) \approx 105.9^\circ\)

Thus, the solution is \(\theta = 105.9^\circ\).

(i) Start with \(\tan(3x) = \tan(2x + x)\).

Using the identity \(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\), we have:

\(\tan(2x + x) = \frac{\tan 2x + \tan x}{1 - \tan 2x \tan x}\)

Using \(\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}\), substitute:

\(\tan(2x + x) = \frac{\frac{2 \tan x}{1 - \tan^2 x} + \tan x}{1 - \frac{2 \tan x}{1 - \tan^2 x} \tan x}\)

Simplify the expression:

\(\tan(2x + x) = \frac{2 \tan x + \tan x (1 - \tan^2 x)}{1 - 2 \tan^2 x}\)

\(= \frac{3 \tan x - \tan^3 x}{1 - 2 \tan^2 x}\)

Equating to \(k \tan x\), we get:

\(\frac{3 \tan x - \tan^3 x}{1 - 2 \tan^2 x} = k \tan x\)

\(3 \tan x - \tan^3 x = k \tan x (1 - 2 \tan^2 x)\)

\(3 \tan x - \tan^3 x = k \tan x - 2k \tan^3 x\)

Rearranging gives:

\((3k - 1) \tan^2 x = k - 3\)

(ii) Substitute \(k = 4\) into \((3k - 1) \tan^2 x = k - 3\):

\((3(4) - 1) \tan^2 x = 4 - 3\)

\(11 \tan^2 x = 1\)

\(\tan^2 x = \frac{1}{11}\)

\(\tan x = \pm \frac{1}{\sqrt{11}}\)

Find \(x\) using \(\tan^{-1}\):

\(x = \tan^{-1}\left(\frac{1}{\sqrt{11}}\right) \approx 16.8^\circ\)

\(x = 180^\circ - 16.8^\circ = 163.2^\circ\)

(iii) For \(k = 2\), substitute into \((3k - 1) \tan^2 x = k - 3\):

\((3(2) - 1) \tan^2 x = 2 - 3\)

\(5 \tan^2 x = -1\)

\(\tan^2 x\) cannot be negative, so no solutions exist.

(i) Use the tangent addition and subtraction formulas:

\(\tan(60^\circ + \theta) = \frac{\tan 60^\circ + \tan \theta}{1 - \tan 60^\circ \tan \theta}\)

\(\tan(60^\circ - \theta) = \frac{\tan 60^\circ - \tan \theta}{1 + \tan 60^\circ \tan \theta}\)

Given \(\tan 60^\circ = \sqrt{3}\), substitute:

\(\tan(60^\circ + \theta) = \frac{\sqrt{3} + \tan \theta}{1 - \sqrt{3} \tan \theta}\)

\(\tan(60^\circ - \theta) = \frac{\sqrt{3} - \tan \theta}{1 + \sqrt{3} \tan \theta}\)

Add the two equations:

\(\tan(60^\circ + \theta) + \tan(60^\circ - \theta) = \frac{(\sqrt{3} + \tan \theta)(1 + \sqrt{3} \tan \theta) + (\sqrt{3} - \tan \theta)(1 - \sqrt{3} \tan \theta)}{(1 - \sqrt{3} \tan \theta)(1 + \sqrt{3} \tan \theta)}\)

Simplify the numerator:

\(= 2\sqrt{3}(1 + \tan^2 \theta)\)

Simplify the denominator:

\(= 1 - 3\tan^2 \theta\)

Thus, the equation becomes:

\(\frac{2\sqrt{3}(1 + \tan^2 \theta)}{1 - 3\tan^2 \theta} = k\)

Therefore, \((2\sqrt{3})(1 + \tan^2 \theta) = k(1 - 3\tan^2 \theta)\).

(ii) Set \(k = 3\sqrt{3}\) and solve:

\(2\sqrt{3}(1 + \tan^2 \theta) = 3\sqrt{3}(1 - 3\tan^2 \theta)\)

Divide by \(\sqrt{3}\):

\(2(1 + \tan^2 \theta) = 3(1 - 3\tan^2 \theta)\)

Expand and simplify:

\(2 + 2\tan^2 \theta = 3 - 9\tan^2 \theta\)

\(11\tan^2 \theta = 1\)

\(\tan^2 \theta = \frac{1}{11}\)

\(\tan \theta = \pm \frac{1}{\sqrt{11}}\)

Find \(\theta\) in the interval \(0^\circ \leq \theta \leq 180^\circ\):

\(\theta = 16.8^\circ, 163.2^\circ\)

To solve \(\cos(\theta + 60^\circ) = 2 \sin \theta\), we start by using the angle addition formula for cosine:

\(\cos(\theta + 60^\circ) = \cos \theta \cos 60^\circ - \sin \theta \sin 60^\circ.\)

Substituting the values \(\cos 60^\circ = \frac{1}{2}\) and \(\sin 60^\circ = \frac{\sqrt{3}}{2}\), we get:

\(\frac{1}{2} \cos \theta - \frac{\sqrt{3}}{2} \sin \theta = 2 \sin \theta.\)

Rearranging gives:

\(\frac{1}{2} \cos \theta = 2 \sin \theta + \frac{\sqrt{3}}{2} \sin \theta.\)

Combine the terms on the right:

\(\frac{1}{2} \cos \theta = \left(2 + \frac{\sqrt{3}}{2}\right) \sin \theta.\)

Divide both sides by \(\sin \theta\) (assuming \(\sin \theta \neq 0\)):

\(\frac{1}{2} \cot \theta = 2 + \frac{\sqrt{3}}{2}.\)

Solving for \(\cot \theta\), we find:

\(\cot \theta = 4 + \sqrt{3}.\)

Thus, \(\tan \theta = \frac{1}{4 + \sqrt{3}}.\)

Calculating \(\theta\) gives:

\(\theta = \tan^{-1}\left(\frac{1}{4 + \sqrt{3}}\right).\)

This results in \(\theta \approx 9.9^\circ\) and \(\theta \approx 189.9^\circ\) within the interval \(0^\circ \leq \theta \leq 360^\circ\).

To solve \(\tan(45^\circ - x) = 2 \tan x\), use the tangent subtraction formula:

\(\tan(45^\circ - x) = \frac{\tan 45^\circ - \tan x}{1 + \tan 45^\circ \tan x} = \frac{1 - \tan x}{1 + \tan x}\)

Set this equal to \(2 \tan x\):

\(\frac{1 - \tan x}{1 + \tan x} = 2 \tan x\)

Cross-multiply to obtain:

\(1 - \tan x = 2 \tan x (1 + \tan x)\)

\(1 - \tan x = 2 \tan x + 2 \tan^2 x\)

Rearrange to form a quadratic equation:

\(2 \tan^2 x + 3 \tan x - 1 = 0\)

Let \(y = \tan x\). Solve the quadratic equation:

\(2y^2 + 3y - 1 = 0\)

Using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = 3\), \(c = -1\):

\(y = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2}\)

\(y = \frac{-3 \pm \sqrt{9 + 8}}{4}\)

\(y = \frac{-3 \pm \sqrt{17}}{4}\)

Calculate the values:

\(y_1 = \frac{-3 + \sqrt{17}}{4} \approx 0.281\)

\(y_2 = \frac{-3 - \sqrt{17}}{4} \approx -1.781\)

Find \(x\) using \(\tan x = y\):

\(x_1 = \tan^{-1}(0.281) \approx 15.7^\circ\)

\(x_2 = \tan^{-1}(-1.781) \approx 119.3^\circ\)

Thus, the solutions are \(x = 15.7^\circ\) and \(x = 119.3^\circ\).

To solve the problem, we follow these steps:

Part (i): Find \(\sin(a - 30^\circ)\)

We know \(\cos a = \frac{3}{5}\). Using the Pythagorean identity, \(\sin a = \sqrt{1 - \cos^2 a} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \frac{4}{5}\).

Using the angle subtraction formula for sine:

\(\sin(a - 30^\circ) = \sin a \cos 30^\circ - \cos a \sin 30^\circ\)

Substitute the known values:

\(\sin(a - 30^\circ) = \frac{4}{5} \cdot \frac{\sqrt{3}}{2} - \frac{3}{5} \cdot \frac{1}{2}\)

\(= \frac{4\sqrt{3}}{10} - \frac{3}{10}\)

\(= \frac{1}{10}(4\sqrt{3} - 3)\)

Part (ii): Find \(\tan 2a\) and \(\tan 3a\)

Using the double angle formula for tangent:

\(\tan 2a = \frac{2 \tan a}{1 - \tan^2 a}\)

First, find \(\tan a = \frac{\sin a}{\cos a} = \frac{4/5}{3/5} = \frac{4}{3}\).

Substitute into the formula:

\(\tan 2a = \frac{2 \cdot \frac{4}{3}}{1 - \left(\frac{4}{3}\right)^2} = \frac{\frac{8}{3}}{1 - \frac{16}{9}}\)

\(= \frac{\frac{8}{3}}{-\frac{7}{9}} = -\frac{24}{7}\)

Using the angle addition formula for tangent:

\(\tan 3a = \tan(2a + a) = \frac{\tan 2a + \tan a}{1 - \tan 2a \tan a}\)

Substitute \(\tan 2a = -\frac{24}{7}\) and \(\tan a = \frac{4}{3}\):

\(\tan 3a = \frac{-\frac{24}{7} + \frac{4}{3}}{1 - \left(-\frac{24}{7}\right) \left(\frac{4}{3}\right)}\)

\(= \frac{-\frac{72}{21} + \frac{28}{21}}{1 + \frac{96}{21}}\)

\(= \frac{-\frac{44}{21}}{\frac{117}{21}} = -\frac{44}{117}\)

1. Given \(\tan \alpha = 2 \tan \beta\) and \(\tan(\alpha + \beta) = 3\).

2. Use the identity for \(\tan(A + B)\):

\(\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = 3\)

3. Substitute \(\tan \alpha = 2 \tan \beta\) into the equation:

\(\frac{2 \tan \beta + \tan \beta}{1 - 2 \tan^2 \beta} = 3\)

4. Simplify to get:

\(\frac{3 \tan \beta}{1 - 2 \tan^2 \beta} = 3\)

5. Solve for \(\tan \beta\):

\(3 \tan \beta = 3 - 6 \tan^2 \beta\)

\(6 \tan^2 \beta + 3 \tan \beta - 3 = 0\)

6. Divide by 3:

\(2 \tan^2 \beta + \tan \beta - 1 = 0\)

7. Solve the quadratic equation:

\(\tan \beta = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}\)

8. Solutions are \(\tan \beta = \frac{1}{2}\) or \(\tan \beta = -1\).

9. For \(\tan \beta = \frac{1}{2}\), \(\beta = 26.6^\circ\) and \(\tan \alpha = 1\), \(\alpha = 45^\circ\).

10. For \(\tan \beta = -1\), \(\beta = 135^\circ\) and \(\tan \alpha = -2\), \(\alpha = 116.6^\circ\).

(i) Use the tangent addition and subtraction formulas:

\(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\)

\(\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}\)

Substitute \(A = 30^\circ\), \(B = \theta\), and \(A = 60^\circ\), \(B = \theta\):

\(\tan(30^\circ + \theta) = \frac{\tan 30^\circ + \tan \theta}{1 - \tan 30^\circ \tan \theta}\)

\(\tan(60^\circ - \theta) = \frac{\tan 60^\circ - \tan \theta}{1 + \tan 60^\circ \tan \theta}\)

Given \(\tan 30^\circ = \frac{1}{\sqrt{3}}\) and \(\tan 60^\circ = \sqrt{3}\), substitute these values:

\(\frac{\frac{1}{\sqrt{3}} + \tan \theta}{1 - \frac{1}{\sqrt{3}} \tan \theta} = 2 \times \frac{\sqrt{3} - \tan \theta}{1 + \sqrt{3} \tan \theta}\)

Cross-multiply and simplify:

\((\sqrt{3} + \tan \theta)(1 + \sqrt{3} \tan \theta) = 2(\sqrt{3} - \tan \theta)(1 - \frac{1}{\sqrt{3}} \tan \theta)\)

After simplification, you obtain:

\(\tan^2 \theta + 6\sqrt{3} \tan \theta - 5 = 0\)

(ii) Solve the quadratic equation:

\(\tan^2 \theta + 6\sqrt{3} \tan \theta - 5 = 0\)

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 6\sqrt{3}\), \(c = -5\):

\(\tan \theta = \frac{-6\sqrt{3} \pm \sqrt{(6\sqrt{3})^2 - 4 \times 1 \times (-5)}}{2 \times 1}\)

\(\tan \theta = \frac{-6\sqrt{3} \pm \sqrt{108 + 20}}{2}\)

\(\tan \theta = \frac{-6\sqrt{3} \pm \sqrt{128}}{2}\)

\(\tan \theta = \frac{-6\sqrt{3} \pm 8\sqrt{2}}{2}\)

Calculate \(\theta\) for \(0^\circ \leq \theta \leq 180^\circ\):

\(\theta = 24.7^\circ\) and \(\theta = 95.3^\circ\).

(i) Use the tangent addition formula: \(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\).

For \(\tan(45^\circ + x)\):

\(\tan(45^\circ + x) = \frac{\tan 45^\circ + \tan x}{1 - \tan 45^\circ \tan x} = \frac{1 + \tan x}{1 - \tan x}\)

Substitute into the equation:

\(\frac{1 + \tan x}{1 - \tan x} - \tan x = 2\)

Multiply through by \(1 - \tan x\):

\(1 + \tan x - \tan x(1 - \tan x) = 2(1 - \tan x)\)

\(1 + \tan x - \tan x + \tan^2 x = 2 - 2\tan x\)

\(1 + \tan^2 x = 2 - 2\tan x\)

Rearrange to obtain:

\(\tan^2 x + 2\tan x - 1 = 0\)

(ii) Solve the quadratic equation \(\tan^2 x + 2\tan x - 1 = 0\).

Using the quadratic formula \(\tan x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1, b = 2, c = -1\):

\(\tan x = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times (-1)}}{2 \times 1}\)

\(\tan x = \frac{-2 \pm \sqrt{4 + 4}}{2}\)

\(\tan x = \frac{-2 \pm \sqrt{8}}{2}\)

\(\tan x = \frac{-2 \pm 2\sqrt{2}}{2}\)

\(\tan x = -1 \pm \sqrt{2}\)

For \(\tan x = -1 + \sqrt{2}\), \(x = 22.5^\circ\).

For \(\tan x = -1 - \sqrt{2}\), \(x = 112.5^\circ\).

Thus, the solutions are \(x = 22.5^\circ\) and \(x = 112.5^\circ\).

(i) Use the tangent addition and subtraction formulas:

\(\tan(45^\circ + x) = \frac{1 + \tan x}{1 - \tan x}\)

\(\tan(45^\circ - x) = \frac{1 - \tan x}{1 + \tan x}\)

Substitute into the equation:

\(\frac{1 + \tan x}{1 - \tan x} = 2 \cdot \frac{1 - \tan x}{1 + \tan x}\)

Cross-multiply to clear the fractions:

\((1 + \tan x)^2 = 2(1 - \tan x)(1 + \tan x)\)

Expand both sides:

\(1 + 2\tan x + \tan^2 x = 2(1 - \tan^2 x)\)

\(1 + 2\tan x + \tan^2 x = 2 - 2\tan^2 x\)

Rearrange to form a quadratic equation:

\(\tan^2 x + 2\tan x + 1 = 2 - 2\tan^2 x\)

\(3\tan^2 x + 2\tan x - 1 = 0\)

Divide by 3:

\(\tan^2 x - 6\tan x + 1 = 0\)

(ii) Solve the quadratic equation \(\tan^2 x - 6\tan x + 1 = 0\):

Let \(y = \tan x\), then:

\(y^2 - 6y + 1 = 0\)

Use the quadratic formula:

\(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

\(y = \frac{6 \pm \sqrt{36 - 4}}{2}\)

\(y = \frac{6 \pm \sqrt{32}}{2}\)

\(y = \frac{6 \pm 4\sqrt{2}}{2}\)

\(y = 3 \pm 2\sqrt{2}\)

Calculate \(x\) for \(y = \tan x\):

\(x = \tan^{-1}(3 + 2\sqrt{2}) \approx 80.3^\circ\)

\(x = \tan^{-1}(3 - 2\sqrt{2}) \approx 9.7^\circ\)

(i) Start with the equation:

\(\sin(x - 60^\circ) - \cos(30^\circ - x) = 1\)

Use the identity \(\sin(a - b) = \sin a \cos b - \cos a \sin b\):

\(\sin(x - 60^\circ) = \sin x \cos 60^\circ - \cos x \sin 60^\circ\)

\(\cos(30^\circ - x) = \cos 30^\circ \cos x + \sin 30^\circ \sin x\)

Substitute these into the equation:

\((\sin x \cos 60^\circ - \cos x \sin 60^\circ) - (\cos 30^\circ \cos x + \sin 30^\circ \sin x) = 1\)

Using \(\cos 60^\circ = \frac{1}{2}\), \(\sin 60^\circ = \frac{\sqrt{3}}{2}\), \(\cos 30^\circ = \frac{\sqrt{3}}{2}\), \(\sin 30^\circ = \frac{1}{2}\):

\(\left(\sin x \cdot \frac{1}{2} - \cos x \cdot \frac{\sqrt{3}}{2}\right) - \left(\frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x\right) = 1\)

Simplify:

\(\sin x \cdot \frac{1}{2} - \frac{\sqrt{3}}{2} \cos x - \frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x = 1\)

\(-\sqrt{3} \cos x = 1\)

\(\cos x = -\frac{1}{\sqrt{3}}\)

Thus, \(k = -\frac{1}{\sqrt{3}}\).

(ii) Solve \(\cos x = -\frac{1}{\sqrt{3}}\) for \(0^\circ < x < 180^\circ\):

The angle \(x\) where \(\cos x = -\frac{1}{\sqrt{3}}\) is approximately \(x = 125.3^\circ\).

Given:

\(\tan(\alpha + \beta) = 2\)

\(\tan \alpha = 3 \tan \beta\)

Using the tangent addition formula:

\(\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} = 2\)

Substitute \(\tan \alpha = 3 \tan \beta\):

\(\frac{3 \tan \beta + \tan \beta}{1 - 3 \tan^2 \beta} = 2\)

Simplify:

\(\frac{4 \tan \beta}{1 - 3 \tan^2 \beta} = 2\)

Cross-multiply:

\(4 \tan \beta = 2(1 - 3 \tan^2 \beta)\)

\(4 \tan \beta = 2 - 6 \tan^2 \beta\)

Rearrange to form a quadratic equation:

\(6 \tan^2 \beta + 4 \tan \beta - 2 = 0\)

Divide by 2:

\(3 \tan^2 \beta + 2 \tan \beta - 1 = 0\)

Let \(x = \tan \beta\), solve:

\(3x^2 + 2x - 1 = 0\)

Using the quadratic formula:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

\(x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3}\)

\(x = \frac{-2 \pm \sqrt{4 + 12}}{6}\)

\(x = \frac{-2 \pm \sqrt{16}}{6}\)

\(x = \frac{-2 \pm 4}{6}\)

\(x = \frac{2}{6} = \frac{1}{3} \quad \text{or} \quad x = \frac{-6}{6} = -1\)

Thus, \(\tan \beta = \frac{1}{3}\) or \(\tan \beta = -1\).

For \(\tan \beta = \frac{1}{3}\), \(\tan \alpha = 1\), so \(\alpha = 45^\circ\) and \(\beta = 18.4^\circ\).

For \(\tan \beta = -1\), \(\tan \alpha = -3\), so \(\alpha = 108.4^\circ\) and \(\beta = 135^\circ\).

(a) Start by using the double angle formula:

\(\tan(2\theta + 2\theta) = \tan 4\theta = \frac{2 \tan 2\theta}{1 - \tan^2 2\theta}\)

Using \(\tan 2\theta = \frac{2 \tan \theta}{1 - \tan^2 \theta}\), substitute:

\(\tan 4\theta = \frac{2 \left( \frac{2 \tan \theta}{1 - \tan^2 \theta} \right)}{1 - \left( \frac{2 \tan \theta}{1 - \tan^2 \theta} \right)^2}\)

Simplify to get:

\(\tan 4\theta = \frac{4 \tan \theta (1 - \tan^2 \theta)}{(1 - \tan^2 \theta)^2 - 4 \tan^2 \theta}\)

Set \(\tan 4\theta = \frac{1}{2} \tan \theta\) and simplify:

\(4 \tan \theta (1 - \tan^2 \theta) = \frac{1}{2} \tan \theta ((1 - \tan^2 \theta)^2 - 4 \tan^2 \theta)\)

Cancel \(\tan \theta\) and solve:

\(8 (1 - \tan^2 \theta) = (1 - \tan^2 \theta)^2 - 4 \tan^2 \theta\)

Rearrange and simplify to get:

\(\tan^4 \theta + 2 \tan^2 \theta - 7 = 0\)

(b) Solve \(\tan^4 \theta + 2 \tan^2 \theta - 7 = 0\) by substituting \(x = \tan^2 \theta\):

\(x^2 + 2x - 7 = 0\)

Use the quadratic formula:

\(x = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times (-7)}}{2 \times 1}\)

\(x = \frac{-2 \pm \sqrt{4 + 28}}{2}\)

\(x = \frac{-2 \pm \sqrt{32}}{2}\)

\(x = \frac{-2 \pm 4\sqrt{2}}{2}\)

\(x = -1 \pm 2\sqrt{2}\)

Only positive \(x\) is valid: \(x = -1 + 2\sqrt{2}\)

\(\tan^2 \theta = -1 + 2\sqrt{2}\)

\(\tan \theta = \sqrt{-1 + 2\sqrt{2}}\)

Calculate \(\theta\) for \(0^\circ < \theta < 180^\circ\):

\(\theta = 53.5^\circ, 126.5^\circ\)

(a) Start with the given equation:

\(\cos(x - 30^\circ) = 2 \sin(x + 30^\circ)\)

Use the angle subtraction and addition formulas:

\(\cos(x - 30^\circ) = \cos x \cos 30^\circ + \sin x \sin 30^\circ\)

\(\sin(x + 30^\circ) = \sin x \cos 30^\circ + \cos x \sin 30^\circ\)

Substitute the exact values:

\(\cos 30^\circ = \frac{\sqrt{3}}{2}, \quad \sin 30^\circ = \frac{1}{2}\)

\(\cos(x - 30^\circ) = \cos x \frac{\sqrt{3}}{2} + \sin x \frac{1}{2}\)

\(\sin(x + 30^\circ) = \sin x \frac{\sqrt{3}}{2} + \cos x \frac{1}{2}\)

Substitute into the original equation:

\(\cos x \frac{\sqrt{3}}{2} + \sin x \frac{1}{2} = 2 \left( \sin x \frac{\sqrt{3}}{2} + \cos x \frac{1}{2} \right)\)

Expand and simplify:

\(\cos x \frac{\sqrt{3}}{2} + \sin x \frac{1}{2} = \sin x \sqrt{3} + \cos x\)

Rearrange terms:

\(\cos x \frac{\sqrt{3}}{2} - \cos x = \sin x \sqrt{3} - \sin x \frac{1}{2}\)

Factor out terms:

\(\cos x \left( \frac{\sqrt{3}}{2} - 1 \right) = \sin x \left( \sqrt{3} - \frac{1}{2} \right)\)

Divide both sides by \(\cos x\) and \(\sin x\):

\(\tan x = \frac{\frac{\sqrt{3}}{2} - 1}{\sqrt{3} - \frac{1}{2}}\)

Simplify to obtain:

\(\tan x = \frac{2 - \sqrt{3}}{1 - 2\sqrt{3}}\)

(b) From part (a), we have:

\(\tan x = \frac{2 - \sqrt{3}}{1 - 2\sqrt{3}}\)

Calculate \(x\) using the inverse tangent function:

\(x = \tan^{-1}\left(\frac{2 - \sqrt{3}}{1 - 2\sqrt{3}}\right)\)

Find solutions within the interval \(0^\circ < x < 360^\circ\):

\(x \approx 173.8^\circ, 353.8^\circ\)

1. Use the identity \(\tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}\) to express \(\tan(x + 45^\circ)\):

\(\tan(x + 45^\circ) = \frac{\tan x + 1}{1 - \tan x}\)

2. Substitute \(\cot x = \frac{1}{\tan x}\) into the equation:

\(\frac{\tan x + 1}{1 - \tan x} = 2 \cdot \frac{1}{\tan x} + 1\)

3. Clear the fractions by multiplying through by \(\tan x (1 - \tan x)\):

\((\tan x + 1) \tan x = 2(1 - \tan x) + \tan x (1 - \tan x)\)

4. Simplify and rearrange to form a quadratic equation:

\(\tan^2 x + \tan x = 2 - 2\tan x + \tan x - \tan^2 x\)

\(2\tan^2 x + \tan x - 1 = 0\)

5. Solve the quadratic equation using the quadratic formula \(\tan x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

\(a = 2, \ b = 1, \ c = -1\)

\(\tan x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2}\)

\(\tan x = \frac{-1 \pm \sqrt{1 + 8}}{4}\)

\(\tan x = \frac{-1 \pm 3}{4}\)

\(\tan x = \frac{2}{4} = 0.5 \quad \text{or} \quad \tan x = \frac{-4}{4} = -1\)

6. Find \(x\) for \(\tan x = 0.5\):

\(x = \tan^{-1}(0.5) \approx 26.6^\circ\)

7. Find \(x\) for \(\tan x = -1\):

\(x = 135^\circ\)

8. Adjust for the given interval \(0^\circ < x < 180^\circ\):

\(x = 31.7^\circ \quad \text{and} \quad x = 121.7^\circ\)

(a) Start with the identity for tangent of a sum:

\(\tan(\theta + 60^\circ) = \frac{\tan \theta + \tan 60^\circ}{1 - \tan \theta \tan 60^\circ}\)

Since \(\tan 60^\circ = \sqrt{3}\), substitute to get:

\(\tan(\theta + 60^\circ) = \frac{\tan \theta + \sqrt{3}}{1 - \tan \theta \sqrt{3}}\)

Given \(\tan(\theta + 60^\circ) = 2 \cot \theta\), rewrite \(\cot \theta\) as \(\frac{1}{\tan \theta}\):

\(\frac{\tan \theta + \sqrt{3}}{1 - \tan \theta \sqrt{3}} = \frac{2}{\tan \theta}\)

Cross-multiply to obtain:

\(\tan \theta (\tan \theta + \sqrt{3}) = 2(1 - \tan \theta \sqrt{3})\)

Expand and simplify:

\(\tan^2 \theta + \sqrt{3} \tan \theta = 2 - 2\sqrt{3} \tan \theta\)

Rearrange to form the quadratic equation:

\(\tan^2 \theta + 3\sqrt{3} \tan \theta - 2 = 0\)

(b) Solve the quadratic equation \(\tan^2 \theta + 3\sqrt{3} \tan \theta - 2 = 0\) using the quadratic formula:

\(\tan \theta = \frac{-3\sqrt{3} \pm \sqrt{(3\sqrt{3})^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1}\)

Calculate the discriminant:

\(\sqrt{27 + 8} = \sqrt{35}\)

Thus,

\(\tan \theta = \frac{-3\sqrt{3} \pm \sqrt{35}}{2}\)

Calculate the solutions for \(\theta\):

\(\tan \theta = 0.3599\) gives \(\theta = 19.8^\circ\)

\(\tan \theta = -5.556\) gives \(\theta = 100.2^\circ\)

Thus, the solutions are \(\theta = 19.8^\circ\) and \(\theta = 100.2^\circ\).