Q13/049) Solve: sin^−1(x)+cos^−1(2x)= π/6

We have cos ^-1 x = sin ^-1 ( 1-4x^2)^(1/2)

now given equation

 sin ^-1(x) = π/6 – cos ^-1(2x)

take sin  of both sides

x = sin (π/6) cos cos ^-1(2x) - cos (π/6) sin  cos ^-1(2x)

or x = ½ * 2x – sqrt(3)/2 * (1- 4x^2)^(1/2) = x - sqrt(3)/2 * (1- 4x^2)^(1/2

or * (1- 4x^2)^(1/2) = 0 => x = +/- ½

One can check by putting the values x = ½ satisfies and -1/2 does not.

So solution = x =1/2