JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Trigonometric Equations

The equation $|\cos x|=\sin x$ requires $\sin x\ge 0$. In one period $[0,2\pi ]$, we have $\sin x\ge 0$ for $x\in [0,\pi ]$. In this interval, $|\cos x|=\sin x$ gives two solutions (one in $(0,\pi /2)$ where $\cos x=\sin x$ at $x=\pi /4$, and one in $(\pi /2,\pi )$ where $-\cos x=\sin x$ at $x=3\pi /4$). So there are 2 solutions per period of $2\pi$. Over the interval $[-4\pi ,4\pi ]$ (length $8\pi$, covering 4 periods), we get $4\times 2=8$ solutions.