Question

If sin^-1(1-x)-2 sin^-1x = \(\frac { π }{ 2 }\), then x is equal to
(A) 0, \(\frac { 1 }{ 2 }\)
(B) 1, \(\frac { 1 }{ 2 }\)
(C) 0
(D) \(\frac { 1 }{ 2 }\)

Answer 1

sin^-1(1-x)-2 sin^-1x = \(\frac { π }{ 2 }\)
Putting \(\frac { π }{ 2 }\) = sin^-1(1 – x) + cos^-1(1 – x),
sin^-1(1 – x) – 2 sin^-1 = sin^-1(1 – x) + cos^-1(1 – x)
⇒ – 2 sin^-1x = cos^-1(1 – x)
Let sin^-1x = α. ∴ sin a = α
∴ – 2 sin^-1x = – 2α = cos^-1(1 – x)
or cos 2α = 1 – x [∵ cos (-θ) = cos θ]
∴ 1 – 2sin²α = (1 – x)
Putting sin α = x, we get
1 – 2x² = 1 – x
or 2x² – x = 0
⇒ x(2x – 1) = 0. ∴ x = 0, \(\frac { 1 }{ 2 }\)
But x = \(\frac { 1 }{ 2 }\) does not satisfy the equation. ∴ x = 0.