## University Calculus: Early Transcendentals (3rd Edition)

$$\theta=\Big[\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\Big]$$
$$\sin^2\theta=\cos^2\theta$$ where $0\le\theta\le2\pi$ *Recall the identity: $\sin^2\theta+\cos^2\theta=1$ So, $\sin^2\theta=1-\cos^2\theta$ That means, $$1-\cos^2\theta=\cos^2\theta$$ $$2\cos^2\theta=1$$ $$\cos^2\theta=\frac{1}{2}$$ Take the square root here: $$\cos\theta=\pm\frac{1}{\sqrt2}=\pm\frac{\sqrt2}{2}$$ - For $\cos\theta=\frac{\sqrt2}{2}$: As $\theta\in[0,2\pi]$, which is the whole unit circle, there are 2 points where $\cos\theta=\frac{\sqrt2}{2}$, which are $\theta=\frac{\pi}{4}$ and $\theta=\frac{7\pi}{4}$. - For $\cos\theta=-\frac{\sqrt2}{2}$: As $\theta\in[0,2\pi]$, which is the whole unit circle, there are also 2 points where $\cos\theta=-\frac{\sqrt2}{2}$, which are $\theta=\frac{3\pi}{4}$ and $\theta=\frac{5\pi}{4}$. Therefore, overall, the answer to this exercise is $$\theta=\Big[\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\Big]$$