Given:
[tex]2\sin \theta-\sqrt[]{2}=0[/tex]
Solve the equation,
[tex]\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=\frac{\pi}{4}+2n\pi,\theta=\frac{3\pi}{4}+2n\pi \\ \theta=\frac{\pi}{4},\frac{3\pi}{4}\in\lbrack0,2\pi) \end{gathered}[/tex]
Answer:
[tex]\theta=\frac{\pi}{4},\frac{3\pi}{4}\in\lbrack0,2\pi)[/tex]
