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# If e

Question:

If $e^{\sin x}-e^{-\sin x}-4=0$, then $x=$

(a) 0

(b) $\sin ^{-1}\left\{\log _{\rho}(2-\sqrt{5})\right\}$

(c) 1

(d) none of these

Solution:

(d) none of these

Given equation: $e^{\sin x}-e^{-\sin x}-4=0$

Let:

$e^{\sin x}=y$

Now,

$y-y^{-1}-4=0$

$\Rightarrow y^{2}-4 y-1=0$

$\therefore y=\frac{4 \pm \sqrt{16+4}}{2}$

$\Rightarrow y=\frac{4 \pm \sqrt{20}}{2}$

$\Rightarrow y=\frac{4 \pm 2 \sqrt{5}}{2}=2 \pm \sqrt{5}$

And,

$y=e^{\sin x}$

$y=e^{\sin x}$

$\Rightarrow e^{\sin x}=2 \pm \sqrt{5}$

Taking log on both sides, we get:

$\sin x=\log _{e}(2 \pm \sqrt{5})$

$\Rightarrow \sin x=\log _{e}(2+\sqrt{5})$ or $\sin x=\log _{e}(2-\sqrt{5})$

$\Rightarrow \sin x=\log _{e}(4.24)$ or $\sin x=\log _{e}(-0.24)$

$\log _{e}(4.24)>1$ and $\sin x$ cannot be greater than 1 .

In the other case, the $\log$ of negative term occur $s$, which is not defined.