JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Trigonometric Equations

Write the equation as $2\tan x=\frac{\pi }{2}-x$. The right side falls strictly from $\frac{\pi }{2}$ at $x=0$ to $-\frac{3\pi }{2}$ at $x=2\pi$, while $2\tan x$ rises strictly on each of the three pieces $\left[0,\frac{\pi }{2}\right)$, $\left(\frac{\pi }{2},\frac{3\pi }{2}\right)$ and $\left(\frac{3\pi }{2},2\pi \right]$ on which it is defined. A strictly rising curve can cross a strictly falling one at most once on each piece.
On $\left[0,\frac{\pi }{2}\right)$ the left side starts at $0$, below $\frac{\pi }{2}$, and grows without bound, so there is one crossing. On $\left(\frac{\pi }{2},\frac{3\pi }{2}\right)$ the left side sweeps from $-\infty$ to $+\infty$ while the right side stays between $-\pi$ and $0$, so there is one crossing. On $\left(\frac{3\pi }{2},2\pi \right]$ the left side climbs from $-\infty$ to $0$ while the right side is at most $-\pi$, so there is one crossing.
The equation has $3$ solutions in $[0,2\pi ]$.