# 4arcsinx + arccosx =pi then x=??

## $4 \arcsin x + \arccos x = \pi$

Jun 30, 2018

$x = \frac{1}{2}$.

#### Explanation:

We know that, $\arcsin x + \arccos x = \frac{\pi}{2}$.

$\therefore \arccos x = \frac{\pi}{2} - \arcsin x$.

Subst.ing in the given eqn. we get,

$4 \arcsin x + \left(\frac{\pi}{2} - \arcsin x\right) = \pi$.

$\therefore 3 \arcsin x = \pi - \frac{\pi}{2} = \frac{\pi}{2}$.

$\therefore \arcsin x = \frac{\pi}{6}$.

$\therefore x = \sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$.

This also satisfy the given eqn.

$\therefore \text{The soln. is } x = \frac{1}{2}$.

Jun 30, 2018

#### Explanation:

First note that $\rightarrow {\sin}^{- 1} x + {\cos}^{- 1} x = \frac{\pi}{2}$

Now given equation is,

$\rightarrow 4 {\sin}^{- 1} x + {\cos}^{- 1} x = \pi$

$\rightarrow 3 {\sin}^{- 1} x + {\sin}^{- 1} x + {\cos}^{- 1} x = \pi$

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

$\rightarrow 3 {\sin}^{- 1} x = \pi - \frac{\pi}{2} = \frac{\pi}{2}$

$\rightarrow {\sin}^{- 1} x = \frac{\pi}{6}$

$\rightarrow x = \sin \left(\frac{\pi}{6}\right) = \frac{1}{2}$