Question

Given that 0 ≤ a < 2π, which of the following gives all solutions to the equation 2cos(2a) + 4cos a - 6 = 0


A. a= 0

B. a = π

C. a = 0 and π

D. a = 0 and 2π

Answer 1

$0 \leq \theta \leq 2\pi;\\ 2\cos(2\theta) + 4\cos(\theta) - 6 = 0|:2\\\cos(2\theta) + 2\cos(\theta) - 3 = 0\\2\cos^2{\theta} - 1 + 2\cos(\theta) - 3 = 0\\2\cos^2{\theta}  + 2\cos(\theta) - 4 = 0$

$Let \ t = \cos(\theta)'\\2t^2 + 2t - 4 = 0 => By \ the \ quadratic formula:\\t = \frac{-2 \pm \sqrt{4 - 4(2)(-4)}}{4} = \frac{-2 \pm \sqrt{36}}{4} = \frac{-2 \pm 6}{4} = \frac{-1 \pm 3}{2} => t \in \{1; -2\}\\\\\cos(\theta) = 1 => \theta \in \{0, \pi\} => the \ answer \ is \ C$