This equation can be rewritten as:
cos^2(x) + 4cos(x) + 3 = 0
Now, let's substitute cos(x) = y:
y^2 + 4y + 3 = 0
This is a quadratic equation that can be factored as:
(y + 3)(y + 1) = 0
So, the solutions to this equation are y = -3 and y = -1.
Now substitute back cos(x) for y:
cos(x) = -3 or cos(x) = -1
However, the range of the cosine function is -1 ≤ cos(x) ≤ 1. Therefore, cos(x) = -3 is not a valid solution.
Therefore, the only valid solution is cos(x) = -1.
This means that x = π or x = 2π.
So, the solutions are x = π and x = 2π.
This equation can be rewritten as:
cos^2(x) + 4cos(x) + 3 = 0
Now, let's substitute cos(x) = y:
y^2 + 4y + 3 = 0
This is a quadratic equation that can be factored as:
(y + 3)(y + 1) = 0
So, the solutions to this equation are y = -3 and y = -1.
Now substitute back cos(x) for y:
cos(x) = -3 or cos(x) = -1
However, the range of the cosine function is -1 ≤ cos(x) ≤ 1. Therefore, cos(x) = -3 is not a valid solution.
Therefore, the only valid solution is cos(x) = -1.
This means that x = π or x = 2π.
So, the solutions are x = π and x = 2π.