Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation cos ⁻¹(x)-2 sin ⁻¹(x)=cos ⁻¹(2 x) is equal to :

Solution

Maths sir · 4 steps

Write everything in terms of sin⁻¹ and the equation becomes 3 sin⁻¹ x = sin⁻¹ 2x.

1. Domain first. cos⁻¹(2x) needs -1 ≤ 2x ≤ 1, so any solution must lie in -1/2 ≤ x ≤ 1/2.
2. Convert to sine. Use cos⁻¹ t = π/2 - sin⁻¹ t on both sides. The left side becomes π/2 - 3 sin⁻¹ x and the right side becomes π/2 - sin⁻¹(2x), so 3 sin⁻¹ x = sin⁻¹(2x).
3. Take sine. Let θ = sin⁻¹ x. Then θ lies in [-π/6, π/6], so 3θ lies in [-π/2, π/2]. Taking sine therefore loses nothing, and the equation is exactly sin 3θ = 2 sin θ.

   3x - 4x³ = 2x ⇒ x(1 - 4x²) = 0 ⇒ x = 0, ±1/2

   x	cos⁻¹ x - 2 sin⁻¹ x	cos⁻¹(2x)
   0	π/2 - 0 = π/2	π/2
   1/2	π/3 - π/3 = 0	cos⁻¹ 1 = 0
   -1/2	2π/3 + π/3 = π	cos⁻¹(-1) = π

4. So. All three values satisfy the original equation, so the sum of the solutions is 0 + 1/2 - 1/2 = 0. The answer is (a).

The idea to keep. Mixed inverse trig equations: convert everything to one function using sin⁻¹ t + cos⁻¹ t = π/2. Before taking sine, check the range of the angle, then verify the endpoint roots.

Answer from the NTA final answer key. Solution by PrepWiser.