Question

1. algebraically solve, using exact value(s), the trigonometric equation: $4\cos^2 x = 3$, $-2\pi \leq x \leq 0$. (2 marks)

solution(s):

Explanation:

Step1: Isolate $\cos^2 x$

Divide both sides by 4: $\cos^2 x = \frac{3}{4}$

Step2: Solve for $\cos x$

Take square roots: $\cos x = \pm \frac{\sqrt{3}}{2}$

Step3: Find reference angle

$\cos \theta = \frac{\sqrt{3}}{2}$ when $\theta = \frac{\pi}{6}$ (reference angle)

Step4: Find $x$ in $[-2\pi,0]$ for $\cos x=\frac{\sqrt{3}}{2}$

Coterminal angles: $-\frac{\pi}{6}, -\frac{11\pi}{6}$

Step5: Find $x$ in $[-2\pi,0]$ for $\cos x=-\frac{\sqrt{3}}{2}$

Coterminal angles: $-\frac{5\pi}{6}, -\frac{7\pi}{6}$

Answer:

$-\frac{11\pi}{6}, -\frac{7\pi}{6}, -\frac{5\pi}{6}, -\frac{\pi}{6}$