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

[JEE Main 2024, 31 Jan (Shift 2)]

Question

The number of solutions, of the equation ${e}^{\sin x}-2{e}^{-\sin x}=2$, is : [JEE Main 2024, 31 Jan (Shift 2)]

YESPYQ · Accepted Answer

Correct answer: d) 0. Let $t=e^{\sin x}>0$ $t-\dfrac{2}{t}=2$ $t^2-2t-2=0$ Since $t>0$ $t=1+\sqrt{3}$ $\sin x=\ln(1+\sqrt{3})$ $\ln(1+\sqrt{3})>1$ $-1\le \sin x\le 1$ the equation has no real solution

The number of solutions, of the equation \({e}^{\sin x}-2{e}^{-\sin x}=2\), is : [JEE Main 2024, 31 Jan (Shift 2)]

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