Question

Find all solutions of the equation in the interval [0, 2 tt).

Answer 1

Given:

`2\sin \theta-\sqrt[]{2}=0`

Solve the equation,

`\begin{gathered} 2\sin \theta-\sqrt[]{2}=0 \\ 2\sin \theta=\sqrt[]{2} \\ \sin \theta=\frac{\sqrt[]{2}}{2} \\ \sin \theta=\frac{1}{\sqrt[]{2}} \\ \theta=\sin ^(-1)(\frac{1}{\sqrt[]{2}}) \\ \theta=(\pi)/(4)+2n\pi,\theta=(3\pi)/(4)+2n\pi \\ \theta=(\pi)/(4),(3\pi)/(4)\in\lbrack0,2\pi) \end{gathered}`

Answer:

`\theta=(\pi)/(4),(3\pi)/(4)\in\lbrack0,2\pi)`