Answer
$x =n \pi +(-1)^{n+1} \dfrac{\pi}{6}$ and $x =n \pi + (-1)^n \dfrac{\pi}{2}$
Work Step by Step
Re-arrange the given equation as:
$ 2 \sin^2 x-\sin x=1$
or, $(2\sin x+1)(\sin x-1)=0$
or, $ 2 (\sin x+\dfrac{1}{2})(\sin x-1)=0$
When $ \sin x=\dfrac{-1}{2}$
Therefore, the general solution of $\sin x$ is:
$x =n \pi +(-1)^{n+1} \dfrac{\pi}{6}$
when $\sin x=1$
Therefore, the general solution of $\sin x$ is:
$x =n \pi + (-1)^n \dfrac{\pi}{2}$
