Question

The range of $x$ for which the formula $3{\sin }^{-1}x={\sin }^{-1}\left[x\right(3-4x^2\left)\right]$ holds is :

• A. $\frac{-2}{3}\le x\le \frac{2}{3}$
• B. $\frac{-1}{2}\le x\le \frac{1}{2}$
• C. $\frac{-1}{4}\le x\le \frac{2}{3}$
• D. $\frac{-1}{3}\le x\le 1$

Answer 1

The correct option is B $\frac{-1}{2}\le x\le \frac{1}{2}$

Putting ${\sin }^{-1}x=A$
So, $3A={\sin }^{-1}\left[\sin A\right(3-4{\sin }^{2}A\left)\right]$
$\Rightarrow 3A={\sin }^{-1}[3\sin A-4{\sin }^{3}A]\Rightarrow 3A={\sin }^{-1}\left[\sin 3A\right]\Rightarrow 3A\in [\frac{-\pi }{2},\frac{\pi }{2}]\Rightarrow A\in [\frac{-\pi }{6},\frac{\pi }{6}]\Rightarrow {\sin }^{-1}x\in [\frac{-\pi }{6},\frac{\pi }{6}]\Rightarrow x\in [\frac{-1}{2},\frac{1}{2}]$