## Trigonometry (11th Edition) Clone

The solution set is $$\{\frac{5\pi}{3}+4\pi n,\frac{7\pi}{3}+4\pi n,n\in Z\}$$
$$\cos\frac{x}{2}+\sqrt3=-\cos\frac{x}{2}$$ 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.