Question Number 211492 by swargiya last updated on 11/Sep/24
Answered by BHOOPENDRA last updated on 11/Sep/24
$${sin}^{−\mathrm{1}} {x}\:−\left(\frac{\pi}{\mathrm{2}}−{sin}^{−\mathrm{1}} {x}\right)=\frac{\pi}{\mathrm{6}} \\ $$$$\mathrm{2}{sin}^{−\mathrm{1}} {x}=\frac{\pi}{\mathrm{6}}+\frac{\pi}{\mathrm{2}} \\ $$$${sin}^{−\mathrm{1}} {x}=\frac{\pi}{\mathrm{3}} \\ $$$${x}={sin}\left(\frac{\pi}{\mathrm{3}}\right)=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$