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

Aug 22, 2016

The Soln. is $x = \pm 1$.

#### Explanation:

${\sin}^{-} 1 x = {\sin}^{-} 1 \left(\frac{1}{x}\right)$

rArr sin(sin^-1x)=sin(sin^-1(1/x)

$\Rightarrow x = \frac{1}{x}$

$\Rightarrow {x}^{2} = 1$

$\Rightarrow x = \pm 1$

These roots satisfy the given eqn.

Hence, the Soln. is $x = \pm 1$.

Aug 22, 2016

$\frac{\pi}{2} , \frac{3 \pi}{2}$

#### Explanation:

arcsin x = arcsin (1/x)
The 2 solutions are:
sin x = 1, and $\sin \left(\frac{1}{x}\right) = 1$
sin x = - 1 , and $\sin \left(\frac{1}{x}\right) = - 1$
Any other values of x will make the equations untrue.
a. sin x = 1 --> arc x = pi/2
b. sin x = - 1 --> arc x = (3pi)/2
$\frac{\pi}{2} , \frac{3 \pi}{2}$