Question
Solve the following equation for x:
$2\tan^{-1}(\sin\text{x})=\tan^{-1}(2\sin\text{x}),\text{x}\neq\frac{\pi}{2}.$

Answer

$2\tan^{-1}(\sin\text{x})=\tan^{-1}(2\sec\text{x})$
$\tan^{-1}\Big(\frac{2\sin\text{x}}{1-\sin^{2}\text{x}}\Big)=\tan^{-1}(2\sec\text{x})$ $\Big[\text{Since }2\tan^{-1}\text{x}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big)\Big]$
$\frac{2\sin\text{x}}{\cos^2\text{x}}=2\sec\text{x}$
$\frac{\sin\text{x}}{\cos\text{x}.\cos\text{x}}=\sec\text{x}$
$\tan\text{x}\sec\text{x}=\sec\text{x}$
$\tan\text{x}=1$
$\text{x}=\frac{\pi}{4}$
Hence the value of x is $\frac{\pi}{4}$
Thus, the solution is $\text{x}=\text{n}\pi+\frac{\pi}{4}$

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