If θ₁ and θ₂ be respectively the smallest and the largest values of θ in (0, 2π) - π which satisfy the equation,
2cot²θ - 5/(sin θ) + 4 = 0, then
∫_(θ ₁)^(θ ₂) cos ²3θ dθ is equal to :

Solution

Maths sir · 4 steps

Turn cot² θ into csc² θ - 1 and the equation becomes a quadratic in csc θ.

1. One function only. Use cot² θ = csc² θ - 1 and 1/(sin θ) = csc θ. The equation becomes 2 csc² θ - 5 csc θ + 2 = 0, which factorises as (2 csc θ - 1)(csc θ - 2) = 0.
2. Pick the valid root. csc θ = 1/2 is impossible, because |csc θ| ≥ 1. So csc θ = 2, that is sin θ = 1/2. In (0, 2π) this gives θ = π/6 and 5π/6, so θ₁ = π/6 and θ₂ = 5π/6.
3. Lower the power. Write cos² 3θ = (1 + cos 6θ)/2, so each part integrates directly.

   ∫_(π/6)^(5π/6) (1 + cos 6θ)/2 dθ = 1/2(5π/6 - π/6) + 1/12(sin 5π - sin π)

4. So. The first part is 1/2 × 2π/3 = π/3. The sine part is 1/12(0 - 0) = 0. The integral equals π/3, which is option (c).

The idea to keep. When cot² θ sits next to 1/(sin θ), switch everything to csc θ and solve the quadratic. Throw out any root with |csc θ| < 1.

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