Question

The number of solution(s) of the equation $e^{\sin x}-e^{-\sin x}-4=0$ is

Answer 1

Given, $e^{\sin x}-e^{-\sin x}-4=0$
Putting $e^{\sin x}=t$, we get
$t-\frac{1}{t}-4=0\Rightarrow t^2-4t-1=0\Rightarrow t=\frac{4\pm \sqrt{20}}{2}\Rightarrow t=2\pm \sqrt{5}\Rightarrow e^{\sin x}=2\pm \sqrt{5}$
As $\sin x\in [-1,1]\Rightarrow e^{\sin x}\in [e^{-1},e]$
Now, we know that
$2+\sqrt{5}>e2-\sqrt{5}<e^{-1}$
So, the given equation has no solution.