## Trigonometry (11th Edition) Clone

The solution set is $$\{2n\pi,\frac{\pi}{6}+2n\pi,\frac{5\pi}{6}+2n\pi,\pi+2n\pi, n\in Z\}$$
$$1-\sin x=\cos 2x$$ - 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 $x$. $$x=\{0+2n\pi,\frac{\pi}{6}+2n\pi,\frac{5\pi}{6}+2n\pi,\pi+2n\pi, n\in Z\}$$ $$x=\{2n\pi,\frac{\pi}{6}+2n\pi,\frac{5\pi}{6}+2n\pi,\pi+2n\pi, n\in Z\}$$