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JEE Main 2021
Trigonometry
Trigonometric Ratio and Identites
Hard

Question

The number of solutions, of the equation esinx2esinx=2e^{\sin x}-2 e^{-\sin x}=2, is :

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Solution

Take esinx=t(t>0)e^{\sin x}=t(t>0) t2t=2t22t=2t22t2=0t22t+1=3(t1)2=3t=1±3t=1±1.73t=2.73 or 0.73 (rejected as t>0)esinx=2.73logeesinx=loge2.73sinx=loge2.73>1\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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