Question

Solve for x: `sin^-1(1 - x) - 2sin^-1(x) = π/2`.

Answer 1

Given that,

`sin^-1(1 - x) - 2sin^-1(x) = π/2`   ...(i)

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

⇒ `1 - x = sin(π/2 + 2 sin^-1(x))`

⇒ 1 – x = cos (2 sin^–1 (x))   ...[Because `sin (π/2 + θ) = cos θ`]

⇒ 1 – x cos{cos^–1 (1 – 2x²)}   ...[Because 2 sin^–1x = cos^–1(1 – 2x²)]

⇒ 1 – x = 1 – 2x²

⇒ x = 2x²

⇒ 2x² – x = 0

⇒ x(2x – 1) = 0

⇒ x = 0, x = `1/2`

For x = `1/2`

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

= `sin^-1 (1/2) - 2 sin^-1 (1/2)`

= `-sin^-1 (1/2)`

= `- π/6 ≠` R.H.S.

Clearly, `x = 1/2` is not a solution.

Therefore, x = 0 is the required solution.