JEE Main 2025 Apr 3 Shift 1, Mathematics Q10: Trigonometric Equations

Rewrite the equation as $\tan x=\frac{\pi }{3}-\frac{2x}{3}.$

Solutions are intersections of $y=\tan x$ (with vertical asymptotes at $x=\pm \frac{\pi }{2},\pm \frac{3\pi }{2}$ excluded) and the line $y=\frac{\pi -2x}{3}$ on the interval $[-2\pi ,2\pi ].$

The line has slope $-\frac{2}{3}$ and passes through $\left(\frac{\pi }{2},0\right)$. On $[-2\pi ,2\pi ]$ the tangent curve has five continuous branches (each open interval between consecutive asymptotes / endpoints): roughly $(-2\pi ,-\frac{3\pi }{2}),(-\frac{3\pi }{2},-\frac{\pi }{2}),(-\frac{\pi }{2},\frac{\pi }{2}),(\frac{\pi }{2},\frac{3\pi }{2}),(\frac{3\pi }{2},2\pi ).$

On each branch where $\tan x$ runs over all of $\mathbb{R}$, exactly one intersection with the line occurs. The endpoint branches at the extreme left and right give one intersection each since $\tan x$ still crosses the line value there. Counting carefully yields 5 intersections.

So the equation has 5 solutions.