Question

Find the exact solution of 5 sin^-1 (x - 2pi) = 2 sin^-1 (x - 3pi) ?

Answer 1

I invalidate this equation.

Explanation:

`sin^(-1)` values `in [ - pi/2, pi/2 ]`. and

`x - 2pi in [ - 1, 1 ]rArr x in [ -1 + 2pi, 1 + 2pi ]`

and, likewise,

`x - 3pi in [ - 1, 1 ] rArr x in [ - 1 + 3pi, 1 +3pi ]`

and the intersection is null.

The two inverse sines are incompatible.