Answer
The solution set is
$$\{\frac{\pi}{3}+2\pi n, \frac{2\pi}{3}+2\pi n, \frac{4\pi}{3}+2\pi n,\frac{5\pi}{3}+2\pi n, n\in Z\}$$
Work Step by Step
$$4\cos^2 x-1=0$$
1) Solve the equation over the interval $[0,2\pi)$
$$4\cos^2 x-1=0$$
$$\cos^2x=\frac{1}{4}$$
$$\cos x=\pm\frac{1}{2}$$
- For $\cos x=\frac{1}{2}$
Over the interval $[0,2\pi)$, there are two values of $x$ where $\cos x=\frac{1}{2}$, which are $\frac{\pi}{3}$ and $\frac{5\pi}{3}$.
- For $\cos x=-\frac{1}{2}$
Over the interval $[0,2\pi)$, there are two values of $x$ where $\cos x=-\frac{1}{2}$, which are $\frac{2\pi}{3}$ and $\frac{4\pi}{3}$.
Therefore, overall, $$x=\{\frac{\pi}{3},\frac{2\pi}{3},\frac{4\pi}{3},\frac{5\pi}{3}\}$$
2) Solve the equation for all solutions
- The integer multiples of the period of the cosine function is $2\pi$.
- We apply it to each solution found in step 1. The results are the solution set.
So the solution set is
$$\{\frac{\pi}{3}+2\pi n, \frac{2\pi}{3}+2\pi n, \frac{4\pi}{3}+2\pi n,\frac{5\pi}{3}+2\pi n, n\in Z\}$$
