## Precalculus (6th Edition) Blitzer

$\{ 1.2515, 5.0317 \}$
Step 1. Use the identity $sin^2x=1-cos^2x$ and let $u=cos(x)$; we have $7u=4-2(1-u^2)$ or $2u^2-7u+2=0$ which gives $u=\frac{7\pm\sqrt {33}}{4}$ or $u\approx0.3139$ and $u\approx3.1861$ Step 2. For $cos(x)=u=3.1861$, there is no solution. Step 3. For $cos(x)=u=0.3139$, we can find the reference angle as $x_0=cos^{-1}(0.3139)\approx1.2515$ and all x-values in $[0,2\pi)$ as $x\approx1.2515$ and $x=2\pi-x_0\approx5.0317$ Step 4. The solutions for the original equation in $[0,2\pi)$ are $\{ 1.2515, 5.0317 \}$