(a) Start with the given equation:

\(\cot^2 \theta + 2 \cos 2\theta = 4\)

Use the identity \(\cot^2 \theta = \csc^2 \theta - 1\):

\(\csc^2 \theta - 1 + 2 \cos 2\theta = 4\)

Substitute \(\csc^2 \theta = \frac{1}{\sin^2 \theta}\) and \(\cos 2\theta = 1 - 2 \sin^2 \theta\):

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

Simplify:

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

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

Multiply through by \(\sin^2 \theta\):

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

Rearrange to:

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

(b) Solve the quadratic equation:

Let \(x = \sin^2 \theta\), then:

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

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

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

\(x = \frac{-3 \pm \sqrt{9 + 16}}{8}\)

\(x = \frac{-3 \pm 5}{8}\)

\(x = \frac{2}{8} = \frac{1}{4} \quad \text{or} \quad x = \frac{-8}{8} = -1\)

Since \(x = \sin^2 \theta\), \(x = -1\) is not valid. So, \(\sin^2 \theta = \frac{1}{4}\).

\(\sin \theta = \pm \frac{1}{2}\)

\(\theta = 30^\circ, 150^\circ, 210^\circ, 330^\circ\)

To solve \(\cot 2\theta = 2 \tan \theta\), we start by using trigonometric identities:

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

Equating this to \(2 \tan \theta\):

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

Cross-multiply to clear the fraction:

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

Rearrange the equation:

\(1 = 5 \tan^2 \theta\)

\(\tan^2 \theta = \frac{1}{5}\)

Take the square root:

\(\tan \theta = \pm \frac{1}{\sqrt{5}}\)

Calculate \(\theta\) using \(\tan^{-1}\):

\(\theta = \tan^{-1} \left( \frac{1}{\sqrt{5}} \right) \approx 24.1^\circ\)

For the second solution in the range \(0^\circ < \theta < 180^\circ\):

\(\theta = 180^\circ - 24.1^\circ = 155.9^\circ\)

Thus, the solutions are \(\theta = 24.1^\circ\) and \(\theta = 155.9^\circ\).

(i) Start with the identity \(\cos^2 x + \sin^2 x = 1\). Raise both sides to the power of 3:

\((\cos^2 x + \sin^2 x)^3 = 1^3 = 1\).

Expand \((\cos^2 x + \sin^2 x)^3\):

\((\cos^2 x + \sin^2 x)^3 = \cos^6 x + 3\cos^4 x \sin^2 x + 3\cos^2 x \sin^4 x + \sin^6 x\).

Using \(\cos^2 x + \sin^2 x = 1\), substitute \(\cos^2 x = 1 - \sin^2 x\) and \(\sin^2 x = 1 - \cos^2 x\):

\(\cos^6 x + \sin^6 x = 1 - 3\cos^2 x \sin^2 x\).

Using the double angle identity \(\sin 2x = 2 \sin x \cos x\), we have \(\sin^2 2x = 4 \sin^2 x \cos^2 x\):

\(\cos^6 x + \sin^6 x = 1 - \frac{3}{4} \sin^2 2x\).

(ii) Given \(\cos^6 x + \sin^6 x = \frac{2}{3}\), substitute the result from part (i):

\(1 - \frac{3}{4} \sin^2 2x = \frac{2}{3}\).

Solve for \(\sin^2 2x\):

\(\frac{3}{4} \sin^2 2x = 1 - \frac{2}{3} = \frac{1}{3}\).

\(\sin^2 2x = \frac{4}{9}\).

\(\sin 2x = \pm \frac{2}{3}\).

For \(\sin 2x = \frac{2}{3}\), solve for \(2x\):

\(2x = \sin^{-1}\left(\frac{2}{3}\right)\) gives \(2x \approx 41.8^\circ\) or \(2x \approx 138.2^\circ\).

Thus, \(x \approx 20.9^\circ\) or \(x \approx 69.1^\circ\).

For \(\sin 2x = -\frac{2}{3}\), solve for \(2x\):

\(2x = \sin^{-1}\left(-\frac{2}{3}\right)\) gives \(2x \approx 221.8^\circ\) or \(2x \approx 318.2^\circ\).

Thus, \(x \approx 110.9^\circ\) or \(x \approx 159.1^\circ\).

Therefore, the solutions are \(x = 20.9^\circ, 69.1^\circ, 110.9^\circ, 159.1^\circ\).

Step 1: Express \(\cot \theta\) and \(\tan \theta\) in terms of \(\sin \theta\) and \(\cos \theta\).

\(\cot \theta = \frac{\cos \theta}{\sin \theta}, \quad \tan \theta = \frac{\sin \theta}{\cos \theta}\)

Step 2: Substitute into the equation:

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

Step 3: Use the identity \(\sin 2\theta = 2 \sin \theta \cos \theta\):

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

Step 4: Multiply through by \(\sin \theta \cos \theta\) to clear the fraction:

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

Step 5: Use \(\sin^2 \theta = 1 - \cos^2 \theta\):

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

Step 6: Simplify:

\(\cos^2 \theta - 2 + 2 \cos^2 \theta = 2 \cos^2 \theta - 2 \cos^4 \theta\)

Step 7: Rearrange to form a quadratic in \(\cos^2 \theta\):

\(2 \cos^4 \theta + \cos^2 \theta - 2 = 0\)

Step 8: Solve for \(\cos^2 \theta\):

Let \(x = \cos^2 \theta\), then:

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

Step 9: Solve the quadratic equation using the quadratic formula:

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

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

\(x = \frac{-1 \pm \sqrt{1 + 16}}{4} = \frac{-1 \pm \sqrt{17}}{4}\)

Step 10: Find \(\theta\) for \(90^\circ < \theta < 180^\circ\):

Only the positive root is valid for \(\cos^2 \theta\), so:

\(\cos^2 \theta = \frac{-1 + \sqrt{17}}{4}\)

\(\theta = 152.1^\circ\)

To express \(\cot 2\theta\) in terms of \(\tan \theta\), use the identity:

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

Given the equation:

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

Multiply through by \(2\tan \theta\) to clear the fraction:

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

Rearrange to form a quadratic equation:

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

Let \(x = \tan \theta\). The equation becomes:

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

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

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

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

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

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

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

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

For \(\tan \theta = \frac{1}{3}\), \(\theta = \tan^{-1}\left(\frac{1}{3}\right) \approx 18.4^\circ\).

For \(\tan \theta = -1\), \(\theta = 135^\circ\) (since \(\tan 135^\circ = -1\)).

Thus, the solutions are \(\theta = 18.4^\circ\) and \(\theta = 135^\circ\).

To express \(\sec \theta = 3 \cos \theta + \tan \theta\) as a quadratic in \(\sin \theta\), we start by using trigonometric identities:

\(\sec \theta = \frac{1}{\cos \theta}\) and \(\tan \theta = \frac{\sin \theta}{\cos \theta}\).

Substitute these into the equation:

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

Multiply through by \(\cos \theta\) to clear the fractions:

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

Use the identity \(\cos^2 \theta = 1 - \sin^2 \theta\):

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

Expand and rearrange:

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

\(3 \sin^2 \theta - \sin \theta - 2 = 0\)

This is a quadratic equation in \(\sin \theta\). Let \(x = \sin \theta\), then:

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

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

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

\(x = \frac{1 \pm \sqrt{1 + 24}}{6}\)

\(x = \frac{1 \pm 5}{6}\)

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

Since \(x = \sin \theta\), and \(-1 \leq \sin \theta \leq 1\), only \(x = -\frac{2}{3}\) is valid in the interval \(-90^\circ < \theta < 90^\circ\).

Thus, \(\sin \theta = -\frac{2}{3}\).

Find \(\theta\):

\(\theta = \arcsin\left(-\frac{2}{3}\right) \approx -41.8^\circ\)

(i) To prove the identity \(\cos 4\theta - 4\cos 2\theta \equiv 8\sin^4\theta - 3\):

Use double angle formulas:

\(\cos 4\theta = 2\cos^2 2\theta - 1\)

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

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

\(\cos 4\theta = 2(1 - 2\sin^2 \theta)^2 - 1\)

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

\(= 2 - 8\sin^2 \theta + 8\sin^4 \theta - 1\)

\(= 8\sin^4 \theta - 8\sin^2 \theta + 1\)

\(\cos 4\theta - 4\cos 2\theta = 8\sin^4 \theta - 8\sin^2 \theta + 1 - 4(1 - 2\sin^2 \theta)\)

\(= 8\sin^4 \theta - 8\sin^2 \theta + 1 - 4 + 8\sin^2 \theta\)

\(= 8\sin^4 \theta - 3\)

Thus, the identity is proven.

(ii) To solve \(\cos 4\theta = 4\cos 2\theta + 3\):

Using the identity \(\cos 4\theta - 4\cos 2\theta = 8\sin^4 \theta - 3\), set \(8\sin^4 \theta - 3 = 3\).

\(8\sin^4 \theta = 6\)

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

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

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

\(\theta = 60^\circ, 120^\circ, 240^\circ, 300^\circ\)

Adjust for the equation \(\cos 4\theta = 4\cos 2\theta + 3\):

\(\theta = 68.5^\circ, 111.5^\circ, 248.5^\circ, 291.5^\circ\)

1. Start with the given equation:

\(\csc \theta = 3 \sin \theta + \cot \theta\)

2. Use the identities \(\csc \theta = \frac{1}{\sin \theta}\) and \(\cot \theta = \frac{\cos \theta}{\sin \theta}\):

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

3. Multiply through by \(\sin \theta\) to clear the fractions:

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

4. Use the Pythagorean identity \(\sin^2 \theta = 1 - \cos^2 \theta\):

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

5. Simplify and rearrange to form a quadratic equation:

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

\(0 = 3\cos^2 \theta - \cos \theta - 2\)

6. Solve the quadratic equation \(3\cos^2 \theta - \cos \theta - 2 = 0\) using the quadratic formula \(\cos \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -1\), \(c = -2\):

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

\(\cos \theta = \frac{1 \pm \sqrt{1 + 24}}{6}\)

\(\cos \theta = \frac{1 \pm 5}{6}\)

\(\cos \theta = 1 \text{ or } \cos \theta = -\frac{2}{3}\)

7. Since \(\cos \theta = 1\) is not in the interval \(0^\circ < \theta < 180^\circ\), we use \(\cos \theta = -\frac{2}{3}\):

\(\theta = \cos^{-1}\left(-\frac{2}{3}\right) \approx 131.8^\circ\)

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

2. Substitute into the equation: \(\frac{\cot^2 x - 1}{2\cot x} + \cot x = 3\).

3. Multiply through by \(2\cot x\) to clear the fraction: \(\cot^2 x - 1 + 2\cot^2 x = 6\cot x\).

4. Simplify to get: \(3\cot^2 x - 6\cot x - 1 = 0\).

5. Let \(y = \cot x\), then the equation becomes \(3y^2 - 6y - 1 = 0\).

6. Solve the quadratic equation using the quadratic formula: \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3, b = -6, c = -1\).

7. Calculate the discriminant: \(b^2 - 4ac = 36 + 12 = 48\).

8. Solve for \(y\): \(y = \frac{6 \pm \sqrt{48}}{6}\).

9. Simplify to find \(y = 3 \pm \sqrt{3}\).

10. Find \(x\) using \(x = \cot^{-1}(y)\).

11. Calculate \(x\) for \(y = 3 + \sqrt{3}\) and \(y = 3 - \sqrt{3}\) within the interval \(0^\circ < x < 180^\circ\).

12. The solutions are \(x = 24.9^\circ\) and \(x = 98.8^\circ\).

(i) We start with the expression \(\sin 2\alpha \sec \alpha\).

Using the identity \(\sin 2\alpha = 2 \sin \alpha \cos \alpha\), we have:

\(\sin 2\alpha \sec \alpha = 2 \sin \alpha \cos \alpha \cdot \sec \alpha\)

Since \(\sec \alpha = \frac{1}{\cos \alpha}\), substitute to get:

\(2 \sin \alpha \cos \alpha \cdot \frac{1}{\cos \alpha} = 2 \sin \alpha\)

Thus, the simplified expression is \(2 \sin \alpha\).

(ii) Given the equation \(3 \cos 2\beta + 7 \cos \beta = 0\), use the identity \(\cos 2\beta = 2 \cos^2 \beta - 1\):

Substitute to get:

\(3(2 \cos^2 \beta - 1) + 7 \cos \beta = 0\)

Simplify to form a quadratic equation:

\(6 \cos^2 \beta - 3 + 7 \cos \beta = 0\)

\(6 \cos^2 \beta + 7 \cos \beta - 3 = 0\)

Solving this quadratic equation using the quadratic formula \(\cos \beta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 6\), \(b = 7\), \(c = -3\):

\(\cos \beta = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 6 \cdot (-3)}}{2 \cdot 6}\)

\(\cos \beta = \frac{-7 \pm \sqrt{49 + 72}}{12}\)

\(\cos \beta = \frac{-7 \pm \sqrt{121}}{12}\)

\(\cos \beta = \frac{-7 \pm 11}{12}\)

This gives two solutions:

\(\cos \beta = \frac{4}{12} = \frac{1}{3}\)

\(\cos \beta = \frac{-18}{12} = -\frac{3}{2}\)

Since \(\cos \beta\) must be within \([-1, 1]\), the valid solution is \(\cos \beta = \frac{1}{3}\).

To solve \(\tan 2x = 5 \cot x\), we start by using the identity \(\cot x = \frac{1}{\tan x}\).

Thus, the equation becomes:

\(\tan 2x = \frac{5}{\tan x}\)

Using the double angle formula for tangent, \(\tan 2x = \frac{2 \tan x}{1 - \tan^2 x}\), we substitute:

\(\frac{2 \tan x}{1 - \tan^2 x} = \frac{5}{\tan x}\)

Cross-multiplying gives:

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

Expanding and rearranging terms, we get:

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

\(7 \tan^2 x = 5\)

\(\tan^2 x = \frac{5}{7}\)

Taking the square root, we find:

\(\tan x = \pm \sqrt{\frac{5}{7}}\)

Now, solve for \(x\) in the interval \(0^\circ < x < 180^\circ\).

For \(\tan x = \sqrt{\frac{5}{7}}\), we find:

\(x = \tan^{-1}\left(\sqrt{\frac{5}{7}}\right) \approx 40.2^\circ\)

For \(\tan x = -\sqrt{\frac{5}{7}}\), we find:

\(x = 180^\circ - \tan^{-1}\left(\sqrt{\frac{5}{7}}\right) \approx 139.8^\circ\)

Thus, the solutions are \(x = 40.2^\circ\) and \(x = 139.8^\circ\).

To solve the equation \(2 \cos x - \cos \frac{1}{2}x = 1\), we can use trigonometric identities and algebraic manipulation.

1. Use the double angle formula: \(\cos x = 2 \cos^2 \frac{x}{2} - 1\).

2. Substitute \(\cos x\) in the equation:

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

3. Simplify the equation:

\(4 \cos^2 \frac{x}{2} - 2 - \cos \frac{x}{2} = 1\)

4. Rearrange to form a quadratic equation in \(\cos \frac{x}{2}\):

\(4 \cos^2 \frac{x}{2} - \cos \frac{x}{2} - 3 = 0\)

5. Let \(u = \cos \frac{x}{2}\). The equation becomes:

\(4u^2 - u - 3 = 0\)

6. Solve the quadratic equation using the quadratic formula:

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

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

\(u = \frac{1 \pm \sqrt{1 + 48}}{8} = \frac{1 \pm 7}{8}\)

7. The solutions for \(u\) are:

\(u = 1\) and \(u = -\frac{3}{4}\).

8. Solve for \(x\):

For \(u = 1\), \(\cos \frac{x}{2} = 1\), so \(\frac{x}{2} = 0\) or \(x = 0\).

For \(u = -\frac{3}{4}\), \(\cos \frac{x}{2} = -\frac{3}{4}\), solve for \(\frac{x}{2}\):

\(\frac{x}{2} = \cos^{-1}(-\frac{3}{4})\), so \(x = 2 \cos^{-1}(-\frac{3}{4})\).

9. Calculate \(x\) for \(\cos^{-1}(-\frac{3}{4})\):

\(x \approx 4.84\).

Thus, the solutions are \(x = 0\) and \(x \approx 4.84\).

To solve the equation \(\csc 2\theta = \sec \theta + \cot \theta\), we start by expressing all terms in terms of sine and cosine:

\(\csc 2\theta = \frac{1}{\sin 2\theta} = \frac{1}{2\sin \theta \cos \theta}\)

\(\sec \theta = \frac{1}{\cos \theta}\)

\(\cot \theta = \frac{\cos \theta}{\sin \theta}\)

Substitute these into the equation:

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

Multiply through by \(2\sin \theta \cos \theta\) to clear the fractions:

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

Using \(\cos^2 \theta = 1 - \sin^2 \theta\), we have:

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

Simplify:

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

\(2\sin^2 \theta - 2\sin \theta - 1 = 0\)

This is a quadratic equation in \(\sin \theta\). Let \(x = \sin \theta\), then:

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

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

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

\(x = \frac{2 \pm \sqrt{4 + 8}}{4}\)

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

\(x = \frac{2 \pm 2\sqrt{3}}{4}\)

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

Thus, \(\sin \theta = \frac{1 + \sqrt{3}}{2}\) or \(\sin \theta = \frac{1 - \sqrt{3}}{2}\).

For \(\sin \theta = \frac{1 + \sqrt{3}}{2}\), solve for \(\theta\):

\(\theta = 201.5^\circ\).

For \(\sin \theta = \frac{1 - \sqrt{3}}{2}\), solve for \(\theta\):

\(\theta = 338.5^\circ\).

Thus, the solutions are \(\theta = 201.5^\circ\) and \(\theta = 338.5^\circ\).

To solve the equation \(\cos \theta + 4 \cos 2\theta = 3\), we start by using the double angle identity \(\cos 2\theta = 2\cos^2\theta - 1\).

Substitute \(\cos 2\theta\) in the equation:

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

Simplify the equation:

\(\cos \theta + 8\cos^2\theta - 4 = 3\)

Rearrange terms:

\(8\cos^2\theta + \cos \theta - 7 = 0\)

This is a quadratic equation in \(\cos \theta\). Let \(x = \cos \theta\), then:

\(8x^2 + x - 7 = 0\)

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 8\), \(b = 1\), \(c = -7\):

\(x = \frac{-1 \pm \sqrt{1^2 - 4 \times 8 \times (-7)}}{2 \times 8}\)

\(x = \frac{-1 \pm \sqrt{1 + 224}}{16}\)

\(x = \frac{-1 \pm \sqrt{225}}{16}\)

\(x = \frac{-1 \pm 15}{16}\)

So, \(x = \frac{14}{16} = \frac{7}{8}\) or \(x = \frac{-16}{16} = -1\).

Thus, \(\cos \theta = \frac{7}{8}\) or \(\cos \theta = -1\).

For \(\cos \theta = \frac{7}{8}\):

\(\theta = \cos^{-1}\left(\frac{7}{8}\right) \approx 29^\circ\)

For \(\cos \theta = -1\):

\(\theta = 180^\circ\)

Therefore, the solutions are \(\theta = 29^\circ\) and \(\theta = 180^\circ\).

1. Start with the equation \(\sin \theta = 2 \cos 2\theta + 1\).

2. Use the double angle formula for cosine: \(\cos 2\theta = 1 - 2\sin^2 \theta\).

3. Substitute into the equation: \(\sin \theta = 2(1 - 2\sin^2 \theta) + 1\).

4. Simplify: \(\sin \theta = 2 - 4\sin^2 \theta + 1\).

5. Rearrange to form a quadratic equation: \(4\sin^2 \theta + \sin \theta - 3 = 0\).

6. Solve the quadratic equation using the quadratic formula: \(\sin \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4, b = 1, c = -3\).

7. Calculate the discriminant: \(b^2 - 4ac = 1 + 48 = 49\).

8. Solve for \(\sin \theta\):

\(\sin \theta = \frac{-1 \pm 7}{8}\)

9. This gives \(\sin \theta = \frac{3}{4}\) or \(\sin \theta = -1\).

10. For \(\sin \theta = \frac{3}{4}\), find \(\theta\):

\(\theta = \sin^{-1}\left(\frac{3}{4}\right) \approx 48.6^\circ\) and \(180^\circ - 48.6^\circ = 131.4^\circ\).

11. For \(\sin \theta = -1\), \(\theta = 270^\circ\).

12. The solutions in the interval \(0^\circ \leq \theta \leq 360^\circ\) are \(\theta = 48.6^\circ, 131.4^\circ, 270^\circ\).

(i) Start with the left-hand side of the identity:

\(\csc 2\theta + \cot 2\theta = \frac{1}{\sin 2\theta} + \frac{\cos 2\theta}{\sin 2\theta}\)

Combine the terms:

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

Using the double angle identity \(\cos 2\theta = 1 - 2\sin^2 \theta\), substitute:

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

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

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

\(= \frac{2\cos^2 \theta}{2\sin \theta \cos \theta}\)

\(= \frac{\cos \theta}{\sin \theta}\)

\(= \cot \theta\)

Thus, the identity is proven.

(ii) Given \(\csc 2\theta + \cot 2\theta = 2\), use the proven identity:

\(\cot \theta = 2\)

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

Calculate \(\theta\):

\(\theta = 26.6^\circ\)

Since \(\cot \theta\) is periodic with period \(180^\circ\), the other solution is:

\(\theta = 26.6^\circ + 180^\circ = 206.6^\circ\)

Thus, \(\theta = 26.6^\circ, 206.6^\circ\).

To solve the equation \(\tan x \tan 2x = 1\), we start by using the identity for \(\tan 2x\):

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

Substitute this into the equation:

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

Simplify the equation:

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

Cross-multiply to clear the fraction:

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

Rearrange the terms:

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

Divide by 3:

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

Take the square root:

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

Thus, \(x = 30^\circ, 150^\circ\) in the given interval \(0^\circ < x < 180^\circ\).

Part (i):

We need to prove the identity \(\cos 4\theta + 4\cos 2\theta \equiv 8\cos^4 \theta - 3\).

Using the double angle formulas:

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

\(\cos 4\theta = 2\cos^2 2\theta - 1 = 2(2\cos^2 \theta - 1)^2 - 1\)

Expanding \(\cos 4\theta\):

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

\(= 8\cos^4 \theta - 8\cos^2 \theta + 2 - 1\)

\(= 8\cos^4 \theta - 8\cos^2 \theta + 1\)

Now, substitute \(\cos 4\theta\) and \(\cos 2\theta\) into the left-hand side:

\(\cos 4\theta + 4\cos 2\theta = (8\cos^4 \theta - 8\cos^2 \theta + 1) + 4(2\cos^2 \theta - 1)\)

\(= 8\cos^4 \theta - 8\cos^2 \theta + 1 + 8\cos^2 \theta - 4\)

\(= 8\cos^4 \theta - 3\)

Thus, the identity is proven.

Part (ii):

We need to solve \(\cos 4\theta + 4\cos 2\theta = 2\).

Using the identity from part (i):

\(8\cos^4 \theta - 3 = 2\)

\(8\cos^4 \theta = 5\)

\(\cos^4 \theta = \frac{5}{8}\)

\(\cos^2 \theta = \sqrt{\frac{5}{8}}\)

\(\cos \theta = \pm \sqrt{\frac{5}{8}}\)

Calculating \(\theta\) for \(\cos \theta = \sqrt{\frac{5}{8}}\):

\(\theta = \cos^{-1}(\sqrt{\frac{5}{8}}) \approx 27.3^\circ\)

Other solutions in the range \(0^\circ \leq \theta \leq 360^\circ\) are:

\(360^\circ - 27.3^\circ = 332.8^\circ\)

For \(\cos \theta = -\sqrt{\frac{5}{8}}\):

\(\theta = 180^\circ - 27.3^\circ = 152.8^\circ\)

\(180^\circ + 27.3^\circ = 207.2^\circ\)

Thus, \(\theta = 27.3^\circ, 152.8^\circ, 207.2^\circ, 332.8^\circ\).

To solve the equation \(\cos \theta + 3 \cos 2\theta = 2\), we start by using the double angle formula for cosine: \(\cos 2\theta = 2\cos^2 \theta - 1\).

Substitute \(\cos 2\theta\) in the equation:

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

Simplify the equation:

\(\cos \theta + 6\cos^2 \theta - 3 = 2\)

Rearrange terms:

\(6\cos^2 \theta + \cos \theta - 5 = 0\)

This is a quadratic equation in terms of \(\cos \theta\). Let \(x = \cos \theta\), then:

\(6x^2 + x - 5 = 0\)

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

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

\(x = \frac{-1 \pm \sqrt{1 + 120}}{12}\)

\(x = \frac{-1 \pm \sqrt{121}}{12}\)

\(x = \frac{-1 \pm 11}{12}\)

\(x = \frac{10}{12} = \frac{5}{6} \quad \text{or} \quad x = \frac{-12}{12} = -1\)

Thus, \(\cos \theta = \frac{5}{6}\) or \(\cos \theta = -1\).

For \(\cos \theta = \frac{5}{6}\):

\(\theta = \cos^{-1} \left( \frac{5}{6} \right) \approx 33.6^\circ\)

For \(\cos \theta = -1\):

\(\theta = 180^\circ\)

Therefore, the solutions are \(\theta = 33.6^\circ\) and \(\theta = 180^\circ\).

(a) Start with the given equation:

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

Using double angle identities, we have:

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

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

Substitute these into the equation:

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

Rearrange terms:

\(\cos^2 \theta + 2 \sin \theta \cos \theta - 3 \sin^2 \theta = 0\)

(b) Factor the equation:

\((\cos \theta - \sin \theta)(\cos \theta + 3 \sin \theta) = 0\)

Set each factor to zero:

\(\cos \theta - \sin \theta = 0\)

\(\tan \theta = 1\)

\(\theta = 45^\circ\)

And:

\(\cos \theta + 3 \sin \theta = 0\)

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

\(\theta \approx 161.6^\circ\)

Thus, the solutions are \(\theta = 45^\circ\) and \(\theta \approx 161.6^\circ\).

(a) Start by expressing \(\cos 4\theta\) and \(\cos 2\theta\) in terms of \(\cos \theta\):

\(\cos 4\theta = 2(\cos 2\theta)^2 - 1 = 2(2\cos^2 \theta - 1)^2 - 1\)

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

Substitute \(\cos 2\theta\) into \(\cos 4\theta\):

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

\(= 8\cos^4 \theta - 8\cos^2 \theta + 2 - 1\)

\(= 8\cos^4 \theta - 8\cos^2 \theta + 1\)

Now, add \(4\cos 2\theta + 3\):

\(\cos 4\theta + 4\cos 2\theta + 3 = 8\cos^4 \theta - 8\cos^2 \theta + 1 + 4(2\cos^2 \theta - 1) + 3\)

\(= 8\cos^4 \theta - 8\cos^2 \theta + 1 + 8\cos^2 \theta - 4 + 3\)

\(= 8\cos^4 \theta\)

Thus, the identity is proven.

(b) Using the identity, solve:

\(8\cos^4 \theta - 3 = 4\)

\(8\cos^4 \theta = 7\)

\(\cos^4 \theta = \frac{7}{8}\)

\(\cos \theta = \pm \sqrt[4]{\frac{7}{8}}\)

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

\(\theta = \cos^{-1}(\sqrt[4]{\frac{7}{8}}) \approx 14.7^\circ\)

\(\theta = 180^\circ - 14.7^\circ = 165.3^\circ\)

Thus, \(\theta = 14.7^\circ, 165.3^\circ\).

To solve the equation \(3 \cos 2\theta = 3 \cos \theta + 2\), we start by using the double-angle identity for cosine: \(\cos 2\theta = 2\cos^2 \theta - 1\).

Substitute this into the equation:

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

Simplify and rearrange:

\(6\cos^2 \theta - 3 = 3 \cos \theta + 2\)

\(6\cos^2 \theta - 3\cos \theta - 5 = 0\)

This is a quadratic equation in terms of \(\cos \theta\). Let \(x = \cos \theta\), then:

\(6x^2 - 3x - 5 = 0\)

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

\(x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 6 \cdot (-5)}}{2 \cdot 6}\)

\(x = \frac{3 \pm \sqrt{9 + 120}}{12}\)

\(x = \frac{3 \pm \sqrt{129}}{12}\)

Calculate the values:

\(x_1 = \frac{3 + \sqrt{129}}{12} \approx 0.694\)

\(x_2 = \frac{3 - \sqrt{129}}{12} \approx -1.194\)

Since \(\cos \theta\) must be between -1 and 1, only \(x_1\) is valid.

Find \(\theta\) using \(\cos^{-1}(x_1)\):

\(\theta = \cos^{-1}(0.694) \approx 134.1^\circ\)

The second solution in the range \([0^\circ, 360^\circ]\) is:

\(\theta = 360^\circ - 134.1^\circ = 225.9^\circ\)

Thus, the solutions are \(\theta = 134.1^\circ\) and \(\theta = 225.9^\circ\).

To solve the equation \(2 \cot 2x + 3 \cot x = 5\), we start by using trigonometric identities.

First, express \(\cot 2x\) using the double angle identity:

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

Substitute \(\cot 2x\) into the equation:

\(2 \left(\frac{1 - \tan^2 x}{2 \tan x}\right) + 3 \cot x = 5\)

Simplify and multiply through by \(\tan x\):

\(1 - \tan^2 x + 3 = 5 \tan x\)

Rearrange to form a quadratic equation:

\(\tan^2 x + 5 \tan x - 4 = 0\)

Let \(y = \tan x\). The equation becomes:

\(y^2 + 5y - 4 = 0\)

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

\(y = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1}\) \(y = \frac{-5 \pm \sqrt{25 + 16}}{2}\) \(y = \frac{-5 \pm \sqrt{41}}{2}\)

Calculate the values of \(y\):

\(y_1 = \frac{-5 + \sqrt{41}}{2} \approx 1.702\) \(y_2 = \frac{-5 - \sqrt{41}}{2} \approx -6.702\)

Since \(y = \tan x\), find \(x\) for \(0^\circ < x < 180^\circ\):

\(x_1 = \tan^{-1}(1.702) \approx 35.1^\circ\) \(x_2 = \tan^{-1}(-6.702) \approx 99.9^\circ\)

Thus, the solutions are \(x = 35.1^\circ\) and \(x = 99.9^\circ\).

(a) Expand \((\cos^2 \theta + \sin^2 \theta)^2\):

\((\cos^2 \theta + \sin^2 \theta)^2 = \cos^4 \theta + 2\cos^2 \theta \sin^2 \theta + \sin^4 \theta\)

Since \(\cos^2 \theta + \sin^2 \theta = 1\), we have:

\(\cos^4 \theta + \sin^4 \theta = 1 - 2\cos^2 \theta \sin^2 \theta\)

Using the double angle identity \(\sin 2\theta = 2\sin \theta \cos \theta\), we get:

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

(b) Solve \(\cos^4 \theta + \sin^4 \theta = \frac{5}{9}\):

Set \(1 - \frac{1}{2} \sin^2 2\theta = \frac{5}{9}\)

\(\frac{1}{2} \sin^2 2\theta = 1 - \frac{5}{9} = \frac{4}{9}\)

\(\sin^2 2\theta = \frac{8}{9}\)

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

Find \(2\theta\):

\(2\theta = \sin^{-1}\left(\frac{2\sqrt{2}}{3}\right)\) or \(2\theta = 180^\circ - \sin^{-1}\left(\frac{2\sqrt{2}}{3}\right)\)

Calculate \(\theta\):

\(\theta = 35.3^\circ, 54.7^\circ, 125.3^\circ, 144.7^\circ\)

(a) Start with the equation:

\(\cot 2\theta + \cot \theta = 2\)

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

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

Combine the terms over a common denominator:

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

Simplify:

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

Multiply through by \(2\tan \theta\):

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

Rearrange into a quadratic equation:

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

(b) Solve the quadratic equation \(\tan^2 \theta + 4\tan \theta - 3 = 0\) using the quadratic formula \(\tan \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1, b = 4, c = -3\):

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

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

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

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

Calculate \(\theta\) for \(\tan \theta = -2 + \sqrt{7}\):

\(\theta = \tan^{-1}(-2 + \sqrt{7}) \approx 0.573\)

Calculate \(\theta\) for \(\tan \theta = -2 - \sqrt{7}\):

\(\theta = \tan^{-1}(-2 - \sqrt{7}) \approx 1.783\)

Thus, the solutions are \(\theta \approx 0.573\) and \(\theta \approx 1.783\).

To solve the equation \(\sin \theta = 3 \cos 2\theta + 2\), we first use the double angle identity for cosine: \(\cos 2\theta = 1 - 2\sin^2 \theta\).

Substitute this into the equation:

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

Simplify the equation:

\(\sin \theta = 3 - 6\sin^2 \theta + 2\)

\(\sin \theta = 5 - 6\sin^2 \theta\)

Rearrange to form a quadratic equation:

\(6\sin^2 \theta + \sin \theta - 5 = 0\)

Let \(x = \sin \theta\). The equation becomes:

\(6x^2 + x - 5 = 0\)

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

\(x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 6 \cdot (-5)}}{2 \cdot 6}\)

\(x = \frac{-1 \pm \sqrt{1 + 120}}{12}\)

\(x = \frac{-1 \pm \sqrt{121}}{12}\)

\(x = \frac{-1 \pm 11}{12}\)

\(x = \frac{10}{12} = \frac{5}{6}\) or \(x = \frac{-12}{12} = -1\)

For \(x = \frac{5}{6}\), \(\theta = \sin^{-1}\left(\frac{5}{6}\right) \approx 56.4^\circ\) and \(180^\circ - 56.4^\circ = 123.6^\circ\).

For \(x = -1\), \(\theta = 270^\circ\).

Thus, the solutions are \(\theta = 56.4^\circ, 123.6^\circ, 270^\circ\).