- A
- B
- C
- D
Step-by-Step Solution
Let θ = 18°, so 2θ = 90° − 3θ and therefore sin2θ = cos3θ. That is 2sinθ·cosθ = 4coθ − 3cosθ = cosθ(1 − 4siθ). Cancelling cosθ (which is not zero) gives 2sinθ = 1 − 4siθ, i.e. with = sinθ. Its positive root is (√5 − 1)/4 ≈ 0.309 = sin18°. The equation has root 0.809 = cos36° instead, and has no real root at all.
Answer: C)