## Precalculus (6th Edition) Blitzer

$0.8959, 2.2456$
Step 1. Simplify the equation and use $cos^2(x)=1-sin^2x$; we have $2-2sin^2x-sin(x)=0$ or $2sin^2x+sin(x)-2=0$ which gives $sin(x)=\frac{-1\pm\sqrt {17}}{4}$ or $sin(x)\approx-1.2808$ and $sin(x)\approx0.7808$ Step 2. For $sin(x)=-1.2808$, there is no solution. Step 3. For $sin(x)=0.7808$, we can find the reference angle as $x_0=sin^{0.7808}\approx0.8959$, thus the solutions in $[0,2\pi)$ are $x\approx0.8959$ and $x=\pi-x_0\approx2.2456$ Step 4. The solutions to the original equation are $0.8959, 2.2456$