(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 \(\theta \approx 1.783\).