Sin On E

$e^{-sin x}$ is positive number, so $e^{sin x}>4$ .

Maximum value of $sinx$ is 1, so maximum value of $e^{sin x}$ is $e$ . $e$ is not greater than 4, so no real solution.

$-1\le \sin { x } \le 1$

$\displaystyle\therefore \quad \cfrac { 1 }{ e } \le { e }^{ \sin { x } }\le e$

But according to the given equation $\displaystyle { e }^{ \sin { x } }=2\pm \sqrt { 5 }$ , which is not possible .

Rearranging the equation gives: $(e^{sin x})^2 - 4e^{sin x} -1 = 0 \Rightarrow\ e^{sin x} = 2 +\sqrt{5} \Rightarrow\ sin x = Ln(2+\sqrt{5}) > 1$ $\therefore\ no \ real \ solutions$

it's easy .....we can't under root minus ?? can we ??