WMC 2018 Senior Problem 35

Depending on which trig identity you prefer, there are several solutions.

• Using the Pythagorean identity $\cos^2 x + \sin^2 x = 1$ $\begin{aligned} f(x) &= \cos^2 x - \sin^2 x + 1 \\ &= \cos^2 x - \sin^2x + \cos^2 x + \sin^2 x \\ &= 2\cos^2 x \end{aligned}$ There are precisely $\boxed{2}$ zeros of $2\cos^2 x$ in the interval $[0,2\pi]$ ( $x=\frac{\pi}{2}$ and $x=\frac{3\pi}{2}$ ).

• Using the Pythagorean identity $\cos(2x) = \cos^2 x - \sin^2 x$ $\begin{aligned} f(x) &= \cos^2 x - \sin^2 x + 1 \\ &= \cos(2x) + 1 \\ \end{aligned}$ Let $\alpha=2x$ . As $x$ ranges over $[0,2\pi]$ , $\alpha$ ranges over $[0,4\pi]$ . There are precisely $\boxed{2}$ points in $[0,4\pi]$ for which $\cos\alpha = -1$ ( $\alpha = \pi$ and $\alpha = 3\pi$ ).