Solve: $\;$ $\sin^{-1} \left(x\right) + \cos^{-1} \left(2x\right) = \dfrac{\pi}{6}$
Let $\;$ $\sin^{-1} \left(x\right) = \alpha$, $\;\;\;$ $\alpha \in \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right]$
Then, $\;$ $x = \sin \alpha$ $\implies$ $\cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - x^2}$
Let $\;$ $\cos^{-1} \left(2x\right) = \beta$, $\;\;\;$ $\beta \in \left[0, \pi\right]$
Then, $\;$ $2 x = \cos \beta$ $\implies$ $\sin \beta = \sqrt{1 - \cos^2 \beta} = \sqrt{1 - 4 x^2}$
Given: $\;$ $\sin^{-1} \left(x\right) + \cos^{-1} \left(2x\right) = \dfrac{\pi}{6}$
i.e. $\;$ $\alpha + \beta = \dfrac{\pi}{6}$
$\therefore \;$ $\sin \left(\alpha + \beta\right) = \sin \left(\dfrac{\pi}{6}\right)$
i.e. $\;$ $\sin \alpha \cos \beta + \cos \alpha \sin \beta = \dfrac{1}{2}$
i.e. $\;$ $ x \times 2x + \sqrt{1 - x^2} \times \sqrt{1 - 4 x^2} = \dfrac{1}{2}$
i.e. $\;$ $\sqrt{1 - 5 x^2 + 4 x^4} = \dfrac{1}{2} - 2 x^2$
i.e. $\;$ $1 - 5 x^2 + 4 x^4 = \left(\dfrac{1}{2} - 2 x^2\right)^2$
i.e. $\;$ $1 - 5 x^2 + 4 x^4 = \dfrac{1}{4} - 2 x^2 + 4 x^4$
i.e. $\;$ $3 x^2 = \dfrac{3}{4}$
i.e. $\;$ $x^2 = \dfrac{1}{4}$ $\implies$ $x = \pm \dfrac{1}{2}$
Verification:
Putting $\;$ $x = \dfrac{1}{2}$ $\;$ in the given equation,
$LHS = \sin^{-1} \left(\dfrac{1}{2}\right) + \cos^{-1} \left(2 \times \dfrac{1}{2}\right) = \dfrac{\pi}{6} + 0 = \dfrac{\pi}{6} = RHS$
Putting $\;$ $x = - \dfrac{1}{2}$ $\;$ in the given equation,
$LHS = \sin^{-1} \left(-\dfrac{1}{2}\right) + \cos^{-1} \left[2 \times \left(-\dfrac{1}{2}\right)\right] = -\dfrac{\pi}{6} + \pi = \dfrac{5 \pi}{6} \neq RHS$
$\therefore \;$ The solution set is $\left\{\dfrac{1}{2} \right\}$