JEE Main · 2025 · 03 Apr 2025 · Shift 1 · Trigonometric Equations

Writing $2x+3\tan x=\pi$ as $\tan x=\frac{\pi }{3}-\frac{2x}{3}$ turns the count into intersections of $y=\tan x$ with a line of slope $-\frac{2}{3}$ on $[-2\pi ,2\pi ]$, excluding the asymptotes $\pm \frac{\pi }{2},\pm \frac{3\pi }{2}$. On each complete branch of $\tan x$ over $\left(-\frac{3\pi }{2},-\frac{\pi }{2}\right)$, $\left(-\frac{\pi }{2},\frac{\pi }{2}\right)$ and $\left(\frac{\pi }{2},\frac{3\pi }{2}\right)$, the tangent sweeps from $-\infty$ to $+\infty$ while the line stays finite, giving one crossing each. On the partial branch $\left[-2\pi ,-\frac{3\pi }{2}\right)$ the tangent rises from $0$ past the line, one crossing, and on $\left(\frac{3\pi }{2},2\pi \right]$ it climbs from $-\infty$ to $0$ and meets the line once. The number of solutions is $5$.