Question

solve the equation for ( 0 leq \theta leq 2pi ). ( sin(\theta)=\frac{sqrt{3}}{2} ). select the pair of solutions.
( pi/3,2pi/3 )
( pi/3,4pi/3 )
( 7pi/6,11pi/6 )
( 5pi/6,7pi/6 )

Explanation:

Step1: Recall the unit - circle values

We know that \(\sin(\theta)=\frac{\sqrt{3}}{2}\) when \(\theta = \frac{\pi}{3}+ 2k\pi\) or \(\theta=\pi-\frac{\pi}{3}+2k\pi\) (\(k\in\mathbb{Z}\)) from the unit - circle definition of the sine function (\(\sin(\alpha)=\sin(\pi - \alpha)\)).

Step2: Find solutions in the interval \(0\leq\theta\leq2\pi\)

For \(k = 0\):

• When \(\theta=\frac{\pi}{3}+2k\pi\), substituting \(k = 0\) gives \(\theta=\frac{\pi}{3}\).
• When \(\theta=\pi-\frac{\pi}{3}+2k\pi=\frac{2\pi}{3}+2k\pi\), substituting \(k = 0\) gives \(\theta=\frac{2\pi}{3}\).

Answer:

\(\frac{\pi}{3},\frac{2\pi}{3}\) (the green box with \(\pi/3,2\pi/3\))