Answer
The solution set is
$$\{\frac{\pi}{3}+2\pi n, \pi+2\pi n,\frac{5\pi}{3}+2\pi n, n\in Z\}$$
Work Step by Step
$$2\cos^2 x+\cos x-1=0$$
1) Solve the equation over the interval $[0,2\pi)$
$$2\cos^2 x+\cos x-1=0$$
$$(2\cos^2 x+2\cos x)+(-\cos x-1)=0$$
$$2\cos x(\cos x+1)-(\cos x+1)=0$$
$$(\cos x+1)(2\cos x-1)=0$$
$$\cos x=-1\hspace{1cm}\text{or}\hspace{1cm}\cos x=\frac{1}{2}$$
- For $\cos x=-1$
Over the interval $[0,2\pi)$, there is one value of $x$ where $\cos x=-1$, which is $\pi$.
- For $\cos x=\frac{1}{2}$
Over the interval $[0,2\pi)$, there is one value of $x$ where $\cos x=\frac{1}{2}$, which are $\frac{\pi}{3}$ and $\frac{5\pi}{3}$.
Therefore, overall, $$x=\{\frac{\pi}{3},\pi,\frac{5\pi}{3}\}$$
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
$$\{\frac{\pi}{3}+2\pi n, \pi+2\pi n,\frac{5\pi}{3}+2\pi n, n\in Z\}$$
