Chapter 7 - Analytic Trigonometry - 7.6 Double-angle and Half-angle Formulas - 7.6 Assess Your Understanding - Page 498: 95

$\frac{\pi}{2},\frac{3\pi}{2}$

Step 1. $f(x)=0 \longrightarrow cos(2x)+sin^2(x)=0 \longrightarrow 1-2sin^2(x)+sin^2(x)=0 \longrightarrow sin^2(x)=1 \longrightarrow sin(x)=\pm1$ Step 2. For $sin(x)=1$, we have $x=2k\pi+\frac{\pi}{2}$ Step 3. For $sin(x)=-1$, we have $x=2k\pi+\frac{3\pi}{2}$ Step 4. Within $[0,2\pi)$, we have $x=\frac{\pi}{2},\frac{3\pi}{2}$