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

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Professor PC Thomas Classes · Answer

(a): We have, $e^{\sin x}-2 e^{-\sin x}=2$ Let $e^{\sin x}=t$\ ∴ (i) becomes $t-\frac{2}{t}=2$\ $\Rightarrow \quad t^{2}-2=2 t \Rightarrow t^{2}-2 t-2=0$\ $\Rightarrow \quad t=\frac{2 \pm \sqrt{4-4(1)(-2)}}{2}$\ $\Rightarrow \quad t=\frac{2 \pm \sqrt{12}}{2}=1 \pm \sqrt{3}$\ Now, $e^{\sin x}=1+\sqrt{3}$ and $e^{\sin x}=1-\sqrt{3}$\ (Not possible)\ $\Rightarrow \quad \sin x=\ln (1+\sqrt{3})>1$ (Not possible)\ $	herefore \quad$ There is no solution.\