I am assuming that we are solving for,
4cos^2 theta +4cos theta +1 = 0
This is a quadratic polynomial of the form;
ax^2 + bx + c = 0
Where x=cos theta. To start, lets get rid of the 4 in the first term. We can divide both sides of the equation by 4 to get;
cos^2 theta + cos theta +1/4 = 0
There are several methods for solving quadratics, but at a glance, I notice that 2 xx 1/2 = 1 and (1/2)^2 = 1/4, which are the last two coefficients in our polynomial. Therefore, we can rewrite the quadratic as;
(cos theta +1/2)^2 = 0
Taking the square root of both sides, we have;
cos theta + 1/2 = 0
So;
cos theta = -1/2
Plugging cos^-1(-1/2) into a calculator, or checking a unit circle will reveal that;
theta = 120^@, 240^@