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