Question

If $4{\sin }^{-1}\left(x\right)+{\cos }^{-1}\left(x\right)=\pi ,$ then $x$ is equal to

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

Answer 1

The correct option is B

$\frac{1}{2}$

Explanation for correct option:

Given,

$4{\sin }^{-1}\left(x\right)+{\cos }^{-1}\left(x\right)=\pi ,$

$\Rightarrow$$\begin{array}{rcl}3{\sin }^{-1}\left(x\right)+\left[{\sin }^{-1}\left(x\right)+{\cos }^{-1}\left(x\right)\right]& =& \pi \end{array}$

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

$\Rightarrow$ $\begin{array}{rcl}3{\sin }^{-1}\left(x\right)& =& \pi –\left(\frac{\pi }{2}\right)\end{array}$

$\Rightarrow$ $\begin{array}{rcl}3{\sin }^{-1}\left(x\right)& =& \frac{\pi }{2}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}{\sin }^{-1}\left(x\right)& =& \frac{\pi }{6}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}x& =& \sin \left(\frac{\pi }{6}\right)\end{array}$

$\begin{array}{rcl}\mathbf{\therefore }\mathit{x}\mathbf{}& \mathbf{=}& \mathbf{}\frac{\mathbf{1}}{\mathbf{2}}\end{array}$

Hence, the correct answer is option B