(a) Start with the equation:

\(5 \cos \theta - \sin \theta \tan \theta + 1 = 0\)

Multiply through by \(\cos \theta\) to eliminate \(\tan \theta\):

\(5 \cos^2 \theta - \sin \theta \frac{\sin \theta}{\cos \theta} \cos \theta + \cos \theta = 0\)

Simplify:

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

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

\(5 \cos^2 \theta - (1 - \cos^2 \theta) + \cos \theta = 0\)

Simplify further:

\(6 \cos^2 \theta + \cos \theta - 1 = 0\)

Thus, \(a = 6\), \(b = 1\), \(c = -1\).

(b) Solve the quadratic equation:

\((3 \cos \theta - 1)(2 \cos \theta + 1) = 0\)

Set each factor to zero:

\(3 \cos \theta - 1 = 0\) gives \(\cos \theta = \frac{1}{3}\)

\(2 \cos \theta + 1 = 0\) gives \(\cos \theta = -\frac{1}{2}\)

Find \(\theta\) for \(\cos \theta = \frac{1}{3}\):

\(\theta \approx 1.23, 5.05\)

Find \(\theta\) for \(\cos \theta = -\frac{1}{2}\):

\(\theta = \frac{2\pi}{3}, \frac{4\pi}{3}\)

Convert to approximate values:

\(\theta \approx 2.09, 4.19\)

Thus, \(\theta = \{1.23, \frac{2\pi}{3}, 2.09, 4.19, 5.05\}\).

(a) Start with the equation:

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

Find a common denominator:

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

Simplify the numerator:

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

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

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

Simplify to get:

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

Thus, the equation is in the form \(a \sin^2 \theta + b \sin \theta + c = 0\) with \(a = 2, b = -2, c = -1\).

(b) Solve the quadratic equation \(2\sin^2 \theta - 2\sin \theta - 1 = 0\) using the quadratic formula:

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

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

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

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

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

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

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

For \(\sin \theta = \frac{1 + \sqrt{3}}{2}\), \(\theta \approx 201.5^\circ\).

For \(\sin \theta = \frac{1 - \sqrt{3}}{2}\), \(\theta \approx 338.5^\circ\).

(a) Let \(u = \sqrt{y}\). Then the equation becomes \(6u + \frac{2}{u} - 7 = 0\).

Multiply through by \(u\) to clear the fraction: \(6u^2 + 2 - 7u = 0\).

Rearrange to form a quadratic: \(6u^2 - 7u + 2 = 0\).

Factor the quadratic: \((2u - 1)(3u - 2) = 0\).

Thus, \(u = \frac{1}{2}\) or \(u = \frac{2}{3}\).

Since \(u = \sqrt{y}\), we have \(\sqrt{y} = \frac{1}{2}\) or \(\sqrt{y} = \frac{2}{3}\).

Therefore, \(y = \left(\frac{1}{2}\right)^2 = \frac{1}{4}\) or \(y = \left(\frac{2}{3}\right)^2 = \frac{4}{9}\).

(b) Use \(\tan x\) in place of \(y\) from part (a).

Thus, \(\tan x = \frac{1}{4}\) or \(\tan x = \frac{4}{9}\).

For \(\tan x = \frac{1}{4}\), \(x = 14.0^\circ\) and \(x = 194.0^\circ\).

For \(\tan x = \frac{4}{9}\), \(x = 24.0^\circ\) and \(x = 204.0^\circ\).

(a) Start with the equation \(4 \cos^4 x + \cos^2 x - 3 = 0\).

Let \(y = \cos^2 x\), then the equation becomes \(4y^2 + y - 3 = 0\).

Factor the quadratic: \((4y - 3)(y + 1) = 0\).

Thus, \(y = \frac{3}{4}\) or \(y = -1\).

Since \(y = \cos^2 x\), \(\cos^2 x = \frac{3}{4}\) (ignore \(y = -1\) as \(\cos^2 x \geq 0\)).

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

Solutions for \(x\) are \(x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}\).

(b) Use the quadratic formula for \(4 \cos^4 x + \cos^2 x - k = 0\).

Let \(y = \cos^2 x\), then \(4y^2 + y - k = 0\).

Using the quadratic formula: \(y = \frac{-1 \pm \sqrt{1 + 16k}}{8}\).

For \(k > 5\), \(1 + 16k > 81\), so \(\sqrt{1 + 16k} > 9\).

Thus, \(y = \frac{-1 \pm \text{something greater than 9}}{8}\).

This implies \(y < 0\) or \(y > 1\), which are not possible for \(\cos^2 x\).

Therefore, no solutions exist for \(k > 5\).

(a) To prove the identity, start by using a common denominator:

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

Simplify the numerator:

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

Factor the numerator:

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

Recognize that \(\sin^2 \theta + \cos^2 \theta = 1\), so \(\cos^2 \theta = 1 - \sin^2 \theta\).

Thus, the expression becomes:

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

Therefore, the identity is proven.

(b) To solve the equation:

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

From part (a), we have:

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

Equating gives:

\(2 \tan^2 \theta = 0\)

Thus, \(\tan^2 \theta = 0\), leading to \(\tan \theta = 0\).

The solutions for \(\tan \theta = 0\) in the interval \(0 < \theta < 2\pi\) are \(\theta = \pi\).