Question
Solve the following:
$\cos^{-1}\text{x}+\sin^{-1}\frac{\text{x}}{2}=\frac{\pi}{6}$

Answer

$\cos^{-1}\text{x}+\sin^{-1}\frac{\text{x}}{2}=\frac{\pi}{6}$
$\Rightarrow\sin^{-1}\frac{\text{x}}{2}=\sin^{-1}\Big(\frac{1}{2}\Big)-\sin^{-1}\Big(\sqrt{1-\text{x}^2}\Big)$
$\Rightarrow\sin^{-1}\frac{\text{x}}{2}=\sin^{-1}\Big[\frac{1}{2}\sqrt{1-1+\text{x}^2}-\sqrt{1-\text{x}^2}\sqrt{1-\frac{1}{4}}\Big]$
$\Rightarrow\frac{\text{x}}{2}=\frac{\text{x}}{2}-\frac{\sqrt3\sqrt{1-\text{x}^2}}{2}$
$\Rightarrow\frac{\sqrt3\sqrt{1-\text{x}^2}}{2}=0$
$\Rightarrow\sqrt{1-\text{x}^2}=0$
$\Rightarrow\text{x}=\pm\frac{1}{2}$

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