Question #f6450

1 Answer
Nov 28, 2017

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^@