Question

please help me with this problem! in the box I put my answer, however, I got it wrong so if possible, please help me correct whatever my mistake was!

Answer 1

Given

$\sin (2\theta)=\frac{1}{2}$$\begin{gathered} 2\theta=sin^{-1}(\frac{1}{2})\text{ } \\ 2\theta=\frac{\pi}{6} \\ divide\text{ both side by 2} \\ \frac{2\theta}{2}=\frac{\pi}{6}\div2 \\ \\ \theta=\frac{\pi}{6}\text{ x }\frac{1}{2} \\ \\ \theta=\frac{\pi}{12}\text{ } \end{gathered}$

But sine is also positive in the second quadrant.

Therefore, the solution to the equation also lie in the second quadrant.

Hence,

$\begin{gathered} \theta=\pi-\frac{\pi}{12} \\ \theta=\frac{12\pi-\pi}{12} \\ \theta=\frac{11\pi}{12} \end{gathered}$

The solution of the equation is

$\theta=\frac{\pi}{12}\text{ and }\theta=\frac{11\pi}{12}$