## Algebra 2 (1st Edition)

$x=\pi +2 \pi n; x= \dfrac{\pi}{3}+2 \pi n ; x= \dfrac{ 5 \pi}{3}+2 \pi n$ and $x=2 \pi n$
Here, we have $2 \sin x \cos x=\sin x$ It has to be noted that $\sin x (2 \cos x-1)=0$ This gives: $\sin x=0 \implies x=\pi, 2 \pi$ The general solution for $\cos (x/2)=\cos (0)$ is $(x/2)=2 n \pi \pm (0)$ $2 \cos x-1=0 \implies \cos x=\dfrac{1}{2}$ This gives: $x=\dfrac{\pi}{3}; \dfrac{ 5\pi}{3}$ Hence, we have $x=\pi +2 \pi n; x= \dfrac{\pi}{3}+2 \pi n ; x= \dfrac{ 5 \pi}{3}+2 \pi n$ and $x=2 \pi n$