JEE Main · 2012 · 29 Apr 2012 · Offline · Trigonometric Equations

Let $t=e^{\sin x}$, so the equation $e^{\sin x}-e^{-\sin x}=4$ becomes $t-\frac{1}{t}=4$, that is $t^2-4t-1=0$, whose roots are $t=2\pm \sqrt{5}$. Now $\sin x$ can only range over $[-1,1]$, so $t=e^{\sin x}$ is trapped in $\left[e^{-1},e\right]\approx [0.37,2.72]$. Neither $2+\sqrt{5}\approx 4.24$ nor $2-\sqrt{5}\approx -0.24$ lies in that interval, so no value of $x$ can satisfy the equation and it has no real roots.