Solve Trigonometric Equations

Using the identity \(\sec(\theta) = \dfrac{1}{\cos(\theta)}\), rewrite the equation:

\[ \cos(\theta) = -\dfrac{\sqrt{3}}{2} \]

Find the reference angle \(\theta_r\) by solving \(\cos(\theta_r) = \dfrac{\sqrt{3}}{2}\):

\[ \theta_r = \dfrac{\pi}{6} \]

Since \(\cos(\theta)\) is negative, the solution lies in Quadrants II or III.

\[ \theta_1 = \pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}, \quad \theta_2 = \pi + \dfrac{\pi}{6} = \dfrac{7\pi}{6} \]

General Solutions: \(\theta = \dfrac{5\pi}{6} + 2n\pi\) and \(\theta = \dfrac{7\pi}{6} + 2n\pi\).

Isolate the sine function: \(\sin(\theta) = -\dfrac{1}{2}\).

Reference angle: \(\theta_r = \dfrac{\pi}{6}\). Sine is negative in Quadrants III and IV.

\[ \theta_1 = \pi + \dfrac{\pi}{6} = \dfrac{7\pi}{6}, \quad \theta_2 = 2\pi - \dfrac{\pi}{6} = \dfrac{11\pi}{6} \]

General Solutions: \(\theta = \dfrac{7\pi}{6} + 2n\pi\) and \(\theta = \dfrac{11\pi}{6} + 2n\pi\).

Let \( \theta = 3x + \frac{\pi}{4} \). Then \( \cos(\theta) = -\frac{1}{\sqrt{2}} \).

Reference angle \( \theta_r = \frac{\pi}{4} \). Solutions for \(\theta\):

\[ \theta_1 = \frac{3\pi}{4} + 2n\pi, \quad \theta_2 = \frac{5\pi}{4} + 2n\pi \]

Substitute back: \( 3x + \frac{\pi}{4} = \frac{3\pi}{4} + 2n\pi \Rightarrow 3x = \frac{\pi}{2} + 2n\pi \Rightarrow x_1 = \frac{\pi}{6} + \frac{2n\pi}{3} \).

And: \( 3x + \frac{\pi}{4} = \frac{5\pi}{4} + 2n\pi \Rightarrow 3x = \pi + 2n\pi \Rightarrow x_2 = \frac{\pi}{3} + \frac{2n\pi}{3} \).

Using \( \sin^2 x = 1 - \cos^2 x \): \[ 2 \cos^{2} x - \cos x - 1 = 0 \]

Factor: \[ (2 \cos x + 1)(\cos x - 1) = 0 \]

1) \( \cos x = -1/2 \Rightarrow x = \frac{2\pi}{3} + 2n\pi, \frac{4\pi}{3} + 2n\pi \).

2) \( \cos x = 1 \Rightarrow x = 2n\pi \).