Question

Solve the following:

`cos^-1x+sin^-1  x/2=π/6`

Answer 1

`cos^-1x+sin^-1 (x/2) = π/6`

`(π/2 − sin^-1x) + sin^-1(x/2) = π/6`

`π/2 − π/6 = sin^-1x−sin^-1(x/2)`

`sin^-1x−sin^-1(x/2) = π/3 = sin^-1(sqrt3/2)`

`sin^-1x = sin^-1(sqrt3/2) + sin^-1(x/2)`

`sin^-1(x) = sin^-1(sqrt3/2 sqrt(1−x^2/4) + x/2 sqrt(1 - 3/4))   ...[sin^-1 x + sin^-1y = sin^-1 [xsqrt(1 - y^2) + ysqrt(1 - x^2)].`

`sin^-1(x) = sin^-1[(sqrt3/2 sqrt(4−x^2)/2) + x/2 . 1/2]`

`x = (sqrt3 sqrt(4 - x^2))/4 + x/4`

`x - x/4 = (sqrt3 sqrt(4 - x^2))/4`

`(3x)/cancel4 =  (sqrt3 sqrt(4 - x^2))/cancel4`

Squaring both the sides

9x² = 3(4 − x²)

3x² = 4 - x²

3x² + x² = 4

4x² = 4

x² = 1

x = ± 1.