Question

If $e^{\sin x}-e^{-\sin x}-4=0$, then x =
(a) 0
(b) ${\sin }^{-1}\left\{{\log }_e\left(2-\sqrt{5}\right)\right\}$
(c) 1
(d) none of these

Answer 1

(d) none of these
Given equation: $e^{\sin x}-e^{-\sin x}-4=0$
Let :
$e^{\sin x}=y$
Now,
$$\begin{gathered} y-y^{-1}-4=0 \\ \Rightarrow y^2-4y-1=0 \end{gathered}$$

∴ $y=\frac{4\pm \sqrt{16+4}}{2}$
$$\begin{gathered} \Rightarrow y=\frac{4\pm \sqrt{20}}{2} \\ \Rightarrow y=\frac{4\pm 2\sqrt{5}}{2}=2\pm \sqrt{5} \end{gathered}$$
And,
$$\begin{gathered} y=e^{\sin x} \\ \Rightarrow e^{\sin x}=2\pm \sqrt{5} \end{gathered}$$
Taking log on both sides, we get:
$\sin x={\log }_e(2\pm \sqrt{5})$
$$\begin{gathered} \Rightarrow \sin x={\log }_e(2+\sqrt{5})\mathrm{or}\sin x={\log }_e(2-\sqrt{5}) \\ \Rightarrow \sin x={\log }_e(4.24)\mathrm{or}\sin x={\log }_e(-0.24) \\ {\log }_{\mathrm{e}}(4.24)>1\mathrm{and}\sin x\mathrm{cannot}\mathrm{be}\mathrm{greater}\mathrm{than}1. \\ \mathrm{In}the\mathrm{other}\mathrm{case},the\log \mathrm{of}\mathrm{negative}\mathrm{term}\mathrm{occur}s,\mathrm{which}\mathrm{is}\mathrm{not}\mathrm{defined}. \end{gathered}$$