Question

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

Answer 1

`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)(sqrt3x)=cos^(-1)(x)=sin^(-1)sqrt(1-x^2)`. So,

`sqrt3x=sqrt(1-x^2)`.

And the only positive solution is `x = 1/2`.