Answer
The solution set is $$\{\frac{5\pi}{3}+4\pi n,\frac{7\pi}{3}+4\pi n,n\in Z\}$$
Work Step by Step
$$\cos\frac{x}{2}+\sqrt3=-\cos\frac{x}{2}$$
1) Normal algebraic operations:
$$2\cos\frac{x}{2}+\sqrt3=0$$
$$\cos\frac{x}{2}=-\frac{\sqrt3}{2}$$
2) Solve the equation for $\frac{x}{2}$ over the interval $[0,2\pi)$
Over the interval $[0,2\pi)$, there are 2 values of $\frac{x}{2}$ with which $\cos\frac{x}{2}=-\frac{\sqrt3}{2}$, which are $\frac{5\pi}{6}$ and $\frac{7\pi}{6}$.
Therefore, $$\frac{x}{2}=\{\frac{5\pi}{6},\frac{7\pi}{6}\}$$
3) Solve for $\frac{x}{2}$, then for $x$ for all solutions
We know that the period of a cosine function is $2\pi$; so we just need to add $2\pi n$ to every solution found in part 2) to have all solutions for $\frac{x}{2}$
$$\frac{x}{2}=\{\frac{5\pi}{6}+2\pi n,\frac{7\pi}{6}+2\pi n,n\in Z\}$$
Now we can solve for $x$:
$$x=\{\frac{5\pi}{3}+4\pi n,\frac{7\pi}{3}+4\pi n,n\in Z\}$$
That is the solution set of this problem.
