Question

If ${\sin }^{-1}x+{\sin }^{-1}2x=\frac{\pi }{3}$ then $x=$

• A. $\frac{\sqrt{3}}{2\sqrt{7}}$
• B. $\sqrt{\frac{3}{7}}$
• C. $0$
• D. $1$

Answer 1

The correct option is A

$\frac{\sqrt{3}}{2\sqrt{7}}$

Explanation for the correct option.

Step 1: Find the relation between angles

Given that, ${\sin }^{-1}x+{\sin }^{-1}2x=\frac{\pi }{3}$.

$\begin{array}{rcl}\theta & =& {\sin }^{-1}x;\varphi ={\sin }^{-1}2x\\ \theta +\varphi & =& \frac{\pi }{3}\\ \varphi & =& \frac{\pi }{3}-\theta ....\left(1\right)\end{array}$

Step 2: Simplify the equation

Taking $\sin$ both sides in equation (1) :

$\begin{array}{rcl}\sin \varphi & =& \sin \left(\frac{\pi }{3}-\theta \right)\\ 2x& =& \sin \frac{\pi }{3}\cos \theta -\cos \frac{\pi }{3}\sin \theta \\ 2x& =& \frac{\sqrt{3}}{2}\sqrt{1-x^2}-\frac{x}{2}\\ \frac{5x}{2}& =& \frac{\sqrt{3}\times \sqrt{1-x^2}}{2}\\ 25x^2& =& 3\left(1-x^2\right)\\ 25x^2-3+3x^2& =& 0\\ 28x^2& =& 3\\ x& =& \pm \frac{\sqrt{3}}{2\sqrt{7}}\end{array}$

Hence, option A is correct.