Let the inverse trigonometric functions take principal values. The number of real solutions of the equation 2 sin ⁻¹ x+3 cos ⁻¹ x=(2 π)/5, is __________.

Solution

Maths sir · 4 steps

Pull out 2(sin⁻¹ x + cos⁻¹ x) and the left side can never drop below π.

1. Use the identity. For every x in [-1, 1], sin⁻¹ x + cos⁻¹ x = π/2. Split 3 cos⁻¹ x as 2 cos⁻¹ x + cos⁻¹ x so that this pair appears twice.

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

2. Solve for the angle. The equation becomes π + cos⁻¹ x = 2π/5, so cos⁻¹ x = -3π/5.
3. Check the range. The principal value of cos⁻¹ x lies in [0, π], so it can never be negative. Put another way, the left side is always at least π, which is bigger than 2π/5.
4. So. No real x satisfies the equation. The number of real solutions is 0.

The common slip: 1. Dropping the minus sign gives cos⁻¹ x = 3π/5, which looks like a valid angle and suggests 1 solution, when the correct count is 0.

The idea to keep. When sin⁻¹ x and cos⁻¹ x appear together, group them into π/2 first. Then check the value you get against the principal range before you count any solutions.

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