The number of elements in the set S = { θ ∈ [ - 4π ,4π ]:3cos ²2θ + 6 cos 2θ - 10cos ²θ + 5 = 0} is __________.

Solution

Maths sir · 4 steps

Turn cos² θ into cos 2θ and the whole thing becomes a quadratic in one variable.

1. One variable. Use 2 cos² θ = 1 + cos 2θ, so 10 cos² θ = 5 + 5 cos 2θ. Put c = cos 2θ: the equation becomes 3c² + 6c - 5 - 5c + 5 = 0.

   3c² + c = 0 ⇒ c(3c+1) = 0 ⇒ cos 2θ = 0 or cos 2θ = -1/3

2. Range of 2θ. Since θ ∈ [-4π, 4π], the angle x = 2θ runs over [-8π, 8π]. That interval has length 16π, which is 8 full periods of cos x.
3. Count each case. In every period, cos x = 0 has 2 solutions and cos x = -1/3 has 2 solutions. At the endpoints x = ± 8π, cos x = 1, so no solution is counted twice. That gives 8 × 2 = 16 solutions for each equation.
4. So. The two cases never overlap, so S has 16 + 16 = 32 elements.

The common slip: 16. Counting over [-4π, 4π] as if it were the range of 2θ gives only 4 periods and a total of 16. The range of 2θ is [-8π, 8π], which doubles the count.

The idea to keep. When the equation mixes cos² θ and cos 2θ, rewrite everything in cos 2θ. Then count solutions over the range of 2θ, not θ, period by period.

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