JEE MAIN - Mathematics (2024 - 31st January Evening Shift - No. 7)

The number of solutions, of the equation $$e^{\sin x}-2 e^{-\sin x}=2$$, is :
0
1
2
more than 2

Explanation

Take $$e^{\sin x}=t(t>0)$$

$$\begin{aligned} & \Rightarrow \mathrm{t}-\frac{2}{\mathrm{t}}=2 \\ & \Rightarrow \frac{\mathrm{t}^2-2}{\mathrm{t}}=2 \\ & \Rightarrow \mathrm{t}^2-2 \mathrm{t}-2=0 \\ & \Rightarrow \mathrm{t}^2-2 \mathrm{t}+1=3 \\ & \Rightarrow(\mathrm{t}-1)^2=3 \\ & \Rightarrow \mathrm{t}=1 \pm \sqrt{3} \\ & \Rightarrow \mathrm{t}=1 \pm 1.73 \\ & \Rightarrow \mathrm{t}=2.73 \text { or }-0.73 \text { (rejected as } \mathrm{t}>0) \\ & \Rightarrow \mathrm{e}^{\sin \mathrm{x}}=2.73 \\ & \Rightarrow \log _{\mathrm{e}} \mathrm{e}^{\sin \mathrm{x}}=\log _{\mathrm{e}} 2.73 \\ & \Rightarrow \sin \mathrm{x}=\log _{\mathrm{e}} 2.73>1 \end{aligned}$$

So no solution.

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