Question

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

Answer 1

The Soln. is `x=+-1`.

Explanation:

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

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

`rArr x=1/x`

`rArr x^2=1`

`rArr x=+-1`

These roots satisfy the given eqn.

Hence, the Soln. is `x=+-1`.

Answer 2

`pi/2, (3pi)/2`

Explanation:

arcsin x = arcsin (1/x)
The 2 solutions are:
sin x = 1, and `sin (1/x) = 1`
sin x = - 1 , and `sin (1/x) = - 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
Answers for (0, 2pi)
`pi/2, (3pi)/2`
Check
arc x = pi/2 --> arcsin (1) = arcsin (1/1)
arc x = (3pi)/2 --> arcsin (-1) = arcsin (1/-1)