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

Solution

Maths sir · 3 steps

Turn every cos² θ into 1-sin² θ and the whole thing becomes a cubic in sin² θ.

1. Factor the cos part. Put s = sin² θ. The cosine terms factor as 3 cos⁴ θ - 5 cos² θ + 2 = (3 cos² θ - 2)(cos² θ - 1). With cos² θ = 1 - s this is (1 - 3s)(-s) = 3s² - s.
2. One variable. The equation is now 3s² - s - 2s³ = 0. Multiply by -1 and factor out s.

   sin² θ (2 sin² θ - 1)(sin² θ - 1) = 0

   Value of sin² θ	θ in [0, 2π]	Count
   0	0, π, 2π	3
   1/2	π/4, 3π/4, 5π/4, 7π/4	4
   1	π/2, 3π/2	2

3. So. No angle appears in two rows, so S has 3 + 4 + 2 = 9 elements. That is option (a).

The idea to keep. When sine and cosine powers are mixed, rewrite everything in sin² θ or cos² θ and factor. On a closed interval like [0, 2π], check both endpoints.

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