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^@#
