Solution

Given the nature of the equation, there are no immediate domain restrictions we need to worry about. Given there are only Cosine functions, I will gather all the terms on one side to see if we can factor: \[ \solve{ 2\cos^2(\theta)&=&1-\cos(\theta)\\ 2\cos^2(\theta)+\cos(\theta)-1&=&0 } \] While more advanced mathematicians may feel perfectly comfortable factoring this expression, I do find it easier for factoring via visual simplification by making the substitution: \(A = \cos(\theta)\): \[\solve{ 2\cos^2(\theta)+\cos(\theta)-1&=&0\\ 2A^2+A-1 &=& 0 } \] Now, we can use the usual rules of factoring. In this case, we will need to use the Group Factoring method since the leading coefficient is a \(2\): we need factors of \(2\times (-1)=-2\) that add up to \(1\), which is conveniently just \(2\) and \(-1\): \[\solve{ 2A^2+A-1 &=& 0\\ 2A^2 +2A - A -1 &=&0\\ 2A(A+1)-1(A+1)&=&0\\ (A+1)(2A-1)&=&0\\ A+1=0&&2A-1=0\\ A=-1&&A=\frac{1}{2} } \] Now, we have found what our temporary variable \(A\) becomes by factoring and applying the Zero Product Principle, but we need to revert to the original problem statement using substitution again: \(A = \cos(\theta)\) \[ \cos(\theta) =-1, \frac{1}{2} \] Under our constraint of \(0\leq \theta\lt2\pi\), this will give us the solutions: \[ \theta = \pi, \frac{\pi}{3}, \frac{5\pi}{3} \] Note that visually, this equation \(2\cos^2(\theta)=1-\cos(\theta)\) is effectively asking us to find the intersection of two separate curves. This is demonstrated in the graph below.