Answer
The solution set is $$\{2n\pi,\frac{\pi}{6}+2n\pi,\frac{5\pi}{6}+2n\pi,\pi+2n\pi, n\in Z\}$$
Work Step by Step
$$1-\sin x=\cos 2x$$
- Recall the identity: $\cos2x=1-2\sin^2x$
$$1-\sin x=1-2\sin^2x$$
$$1-\sin x-1+2\sin^2x=0$$
$$2\sin^2x-\sin x=0$$
$$\sin x(2\sin x-1)=0$$
$$\sin x=0\hspace{1cm}\text{or}\hspace{1cm}\sin x=\frac{1}{2}$$
1) First, we solve the equation over the interval $[0,2\pi)$
Over the interval $[0,2\pi)$, $\{0,\pi\}$ are two values of $x$ where $\sin x=0$, and $\{\frac{\pi}{6},\frac{5\pi}{6}\}$ are two values of $x$ where $\sin x=\frac{1}{2}$
Overall, $$x=\{0,\frac{\pi}{6},\frac{5\pi}{6},\pi\}$$
2) Solve the equation for all solutions
Sine function has period $2\pi$, so we would add $2\pi$ to all solutions found in part 1) for $x$.
$$x=\{0+2n\pi,\frac{\pi}{6}+2n\pi,\frac{5\pi}{6}+2n\pi,\pi+2n\pi, n\in Z\}$$
$$x=\{2n\pi,\frac{\pi}{6}+2n\pi,\frac{5\pi}{6}+2n\pi,\pi+2n\pi, n\in Z\}$$
