Question

If sin^-1(1 - x) - 2 sin^-1x =  `pi/2`, then find the value of x.

Answer 1

sin^-1(1 - x) - 2 sin^-1x =  `pi/2`

∴ sin^-1(1 - x) = `pi/2 + 2 sin^-1 "x"`

∴ 1 - x = sin `(pi/2 + 2 sin^-1"x")`

∴ 1 - x = cos (2 sin^-1x) ....`[ ∵ sin (pi/2 + theta) = cos theta]`

∴ 1 - x = 1 - 2[sin(sin^-1 x)]²   ....[∵ cos 2θ = 1 - 2 sin²θ]

∴ 1 - x = 1 - 2x²

∴ 2x² - x = 0

∴ x(2x - 1) = 0

∴ x = 0  or x = `1/2`

When x = `1/2`

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

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

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

`= - sin^-1 (sin  pi/6)`

`= - pi/6 ne pi/2`

∴ `"x" ne 1/2`

Hence, x = 0.