Question

solve for the angle θ, where 0 ≤ θ ≤ 2π.

$sin^2 \theta = \frac{3}{4}$

$\theta = square$
(use a comma to separate answers as needed. type an exact answer, using π as needed.)

Explanation:

Step1: Take square root of both sides

Given \(\sin^{2}\theta=\frac{3}{4}\), take the square root of both sides: \(\sin\theta = \pm\sqrt{\frac{3}{4}}=\pm\frac{\sqrt{3}}{2}\)

Step2: Solve for \(\theta\) when \(\sin\theta=\frac{\sqrt{3}}{2}\)

We know that \(\sin\theta=\frac{\sqrt{3}}{2}\) when \(\theta = \frac{\pi}{3}\) (in the first quadrant) and \(\theta=\pi - \frac{\pi}{3}=\frac{2\pi}{3}\) (in the second quadrant) within the interval \(0\leq\theta\leq2\pi\).

Step3: Solve for \(\theta\) when \(\sin\theta=-\frac{\sqrt{3}}{2}\)

We know that \(\sin\theta = -\frac{\sqrt{3}}{2}\) when \(\theta=\pi+\frac{\pi}{3}=\frac{4\pi}{3}\) (in the third quadrant) and \(\theta = 2\pi-\frac{\pi}{3}=\frac{5\pi}{3}\) (in the fourth quadrant) within the interval \(0\leq\theta\leq2\pi\).

Answer:

\(\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}\)