Trigonometrical Equations 3 Question 17
18. Show that the equation $e^{\sin x}-e^{-\sin x}-4=0$ has no real solution.
$(1982,2 M)$
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Answer:
Correct Answer: 18. $\frac{\pi}{8}, \cos \frac{\pi}{8} \quad \frac{\pi}{4}, \cos \frac{\pi}{4}-\frac{3 \pi}{8}, \cos \frac{3 \pi}{8}$
Solution:
- Given,
$$ e^{\sin x}-\frac{1}{e^{\sin x}}=4 $$
$$ \begin{aligned} & \Rightarrow \quad\left(e^{\sin x}\right)^{2}-4\left(e^{\sin x}\right)-1=0 \\ & \Rightarrow \quad e^{\sin x}=\frac{4 \pm \sqrt{16+4}}{2}=2 \pm \sqrt{5} \end{aligned} $$
But since, $e \sim 2.72$ and we know, $0<e^{\sin x}<e$ $\therefore \quad e^{\sin x}=2 \pm \sqrt{5}$ is not possible.
Hence, it does not exist any solution.
