Question

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

Answer 1

sin^-1(1 – x) – 2 sin^-1 x = π/2

⇒ sin^-1(1 – x) – 2 sin^-1x

= sin^-1(1 – x) + cos ^-1(1 – x)

⇒ – 2 sin^-1 x

= cos^-1(1 – x)

[∴ sin^-1 (1 – x) + cos^-1 (1 – x) ⇒ π/2]

Let sin^-1x = θ

then sin θ = x

Let sin^-1x = θ

⇒ sin θ = x

∴ – 2θ = cos^-1(1 – x)

⇒ cos(- 2θ) = 1 – x

⇒ cos 2θ = 1 – x …………(i)

Since, we know that

cos 2θ ⇒ 1 – 2 sin² θ

⇒ cos 2θ ⇒ 1 – 2x² …………. (ii)

From equations (i) and (ii), we have

1 – x = 1 – 2x²

⇒ 2x² – x = θ ⇒ x (2x – 1) = 0

⇒ x = 0, x = 1/2.

But x = 1/2 does not satisfies given equation.

∴ x = 0