## Thomas' Calculus 13th Edition

$\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}$
From $sin^2\theta=cos^2\theta$, we have $sin^2\theta=1-sin^2\theta$ and thus $sin^2\theta=\frac{1}{2}$. This leads to $sin\theta=\frac{\sqrt 2}{2}$ and $sin\theta=-\frac{\sqrt 2}{2}$. Within $[0,2\pi]$, the solutions are: $\theta=\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}$