JEE Main · 2024 · 31 Jan 2024 · Shift 2 · Trigonometric Equations

Let $t=e^{\sin x}$, so $t>0$ and, since $\sin x\in [-1,1]$, $t\in \left[\frac{1}{e},e\right]$.

The equation $e^{\sin x}-2e^{-\sin x}=2$ becomes $t-\frac{2}{t}=2$, i.e.

$t^2-2t-2=0\Rightarrow t=\frac{2\pm \sqrt{4+8}}{2}=1\pm \sqrt{3}$.

$t=1-\sqrt{3}<0$ is rejected (as $t>0$).

$t=1+\sqrt{3}\approx 2.732$, but $e\approx 2.718$, so $t=1+\sqrt{3}>e$, which lies outside the allowed range $\left[\frac{1}{e},e\right]$.

Since no valid $t$ exists, no real $x$ satisfies the equation: the number of solutions is $0$.