How do you find the solutions to sin^-1(sqrt3x)=cos^-1x?

Dec 17, 2016

$\frac{1}{2}$

Explanation:

sin^(-1)(sqrt3x) in Q_1 or Q+4) and cos^(-1)x in Q_1 or Q_2.

For parity, both $\in {Q}_{1} \to x > 0$

${\sin}^{- 1} \left(\sqrt{3} x\right) = {\cos}^{- 1} \left(x\right) = {\sin}^{- 1} \sqrt{1 - {x}^{2}}$. So,

$\sqrt{3} x = \sqrt{1 - {x}^{2}}$.

And the only positive solution is $x = \frac{1}{2}$.