If √3 (cos ²x) = (√3 - 1)cos x + 1, the number of solutions of the given equation when x ∈ [ 0,π/2 ] is __________.

Solution

Maths sir · 4 steps

Call cos x a single variable, factor the quadratic, then ask which root cosine can actually reach.

1. Make it quadratic. Put c = cos x and bring everything to one side. You get √3c² - (√3 - 1)c - 1 = 0.

   √3c² - √3c + c - 1 = (√3c + 1)(c - 1) = 0

2. Two roots. So cos x = 1 or cos x = -1/√3.
3. Check the interval. For x ∈ [0, π/2], cos x lies between 0 and 1, so it is never negative. The root -1/√3 is rejected. The value cos x = 1 is reached only at x = 0, which is inside the interval.
4. So. The only solution is x = 0, so the number of solutions is 1.

The common slip: 2. Keeping cos x = -1/√3 as a second root gives 2, but cosine is never negative on [0, π/2].

The idea to keep. For a quadratic in cos x, split the middle term to factor it. Then keep only the roots that cosine can take on the given interval.

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