Question

The equation $e^{\sin x}-e^{-\sin x}-4=0$ has

• A. No solution
• B. Two solutions
• C. Three solutions
• D. None of these

Answer 1

The correct option is A No solution

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

multiply throughout by $e^{sinx}$

$\Rightarrow e^{2\sin x}-4e^{\sin x}-1=0$

this is quadratc equation in $e^{sinx}$

converting from exponential to logarithmic form.

$e^{\sin x}=\frac{4\pm \sqrt{16+4}}{2}=2\pm \sqrt{5}$

$\Rightarrow \sin x=\log (2+\sqrt{5})$ [$\because \log (2-\sqrt{5})$ is not defined ]

Since, $2+\sqrt{5}>e\Rightarrow \log (2+\sqrt{5})>1$

$\Rightarrow \sin x>1$, which is not possible.
Hence, no solution exist.