Question

Solve the equation on the interval 0<=theta<2pi2 sin^2 theta-V3 sin theta = 0

Answer 1

Given the equation

$2\sin ^2\theta-\sqrt{3}\sin \theta=0$

We can rewrite the given equation as:

$2\sin ^{}\theta\sin ^{}\theta-\sqrt{3}\sin \theta=0$

Factoring, we obtain:

$\sin ^{}\theta(2\sin ^{}\theta-\sqrt{3})=0$

Therefore, we can say:

$\begin{gathered} \sin ^{}\theta=0\: \implies\theta=0 \\ \text{Similarly} \\ 2\sin ^{}\theta-\sqrt{3}=0 \\ \implies2\sin ^{}\theta=\sqrt{3} \\ \implies\sin ^{}\theta=\frac{\sqrt{3}}{2} \\ \implies\theta=\sin ^{-1}\frac{\sqrt{3}}{2} \\ \implies\theta=\frac{\pi}{3} \end{gathered}$

Therefore, the solution set in the given interval is:

$\left\lbrace 0,\frac{\pi}{3}\right\rbrace$