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-√(3)\sin \theta=0`

We can rewrite the given equation as:

`2\sin ^{}\theta\sin ^{}\theta-√(3)\sin \theta=0`

Factoring, we obtain:

`\sin ^{}\theta(2\sin ^{}\theta-√(3))=0`

Therefore, we can say:

`\begin{gathered} \sin ^{}\theta=0\: \implies\theta=0 \\ \text{Similarly} \\ 2\sin ^{}\theta-√(3)=0 \\ \implies2\sin ^{}\theta=√(3) \\ \implies\sin ^{}\theta=(√(3))/(2) \\ \implies\theta=\sin ^(-1)(√(3))/(2) \\ \implies\theta=(\pi)/(3) \end{gathered}`

Therefore, the solution set in the given interval is:

`\left\lbrace 0,(\pi)/(3)\right\rbrace`