Question

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

विकल्प

• 0

• \[\sin^{- 1} \left\{ \log_e \left( 2 - \sqrt{5} \right) \right\}\]

   

• 1

• none of these

Answer 1

 none of these
Given equation:
\[e^{\sin x} - e^{- \sin x} - 4 = 0\]
Let :
\[e^{\sin x }= y\]
Now,
\[y - y^{- 1} - 4 = 0\]
\[ \Rightarrow y^2 - 4y - 1 = 0\]

∴ \[y = \frac{4 \pm \sqrt{16 + 4}}{2}\]
\[\Rightarrow y = \frac{4 \pm \sqrt{20}}{2}\]
\[ \Rightarrow y = \frac{4 \pm 2\sqrt{5}}{2} = 2 \pm \sqrt{5}\]
and,
\[y = e^{\sin x} \]
\[ \Rightarrow e^{\sin x} = 2 \pm \sqrt{5}\]
Taking log on both sides, we get:

\[\sin x = \log_e (2 \pm \sqrt{5})\]

\[\Rightarrow \sin x = \log_e ( 2 + \sqrt{5})\text{ or }\sin x = \log_e ( 2 - \sqrt{5})\]
\[ \Rightarrow \sin x = \log_e ( 4 . 24)\text{ or }\sin x = \log_e ( - 0 . 24)\]
\[ \log_e ( 4 . 24) > 1\text{ and }\sin x\text{ cannot be greater than }1 . \]
In the other case, the log of negative term occurs, which is not defined.