Question

Find all solutions in the interval [0, 2π).

Answer 1

First you must know that for remarkable angles: cos (0) = 1, cos (π) = - 1, cos (π / 2) = 0, cos (3π / 2) = 0, cos (2π) = 1. Then, by simple substitution in the given formula, you can find the solutions of x. Which for the interval [0, 2π) are: x = π, x = pi divided by two and x = three pi divided by two.Attached solution.

Answer 2

We will simplify the equation: 
$\cos2x+2\cos x+1=0\\\text{We know the formula } \cos 2x=\cos^2x-\sin^2x\text{ and the formula }\\\cos^2x+\sin^2x=1\\\text{Then the equation becomes :}\\\cos^2x-\sin^2x+2\cos x+cos^2x+\sin^2x=0\\\text{Simplifying we get:}\\2\cos^2x+2\cos x=0\\\text{Factoring:}\\cosx(cosx+1)=0\\\text{Then either }cosx=0\text{ or }cosx+1=0\\\text{From the first equation we get }x=\frac{\pi}{2},x=\frac{3\pi}{2}\\\text{and from the second equation we get } x=\pi$

The correct answer is the last one plus pi.