Answer
$$\theta=\Big[\frac{\pi}{3},\pi,\frac{5\pi}{3}\Big]$$
Work Step by Step
$$\cos2\theta+\cos\theta=0$$
where $0\le\theta\le2\pi$
Apply the Double Angle Formula here for $\cos2\theta$, which states $\cos2\theta=\cos^2\theta-\sin^2\theta$
Yet, we also recall that $\sin^2\theta=1-\cos^2\theta$.
Therefore, $$\cos2\theta=\cos^2\theta-(1-\cos^2\theta)=\cos^2\theta-1+\cos^2\theta=2\cos^2\theta-1$$
That means, $$2\cos^2\theta-1+\cos\theta=0$$
$$2\cos^2\theta+\cos\theta-1=0$$
$$(2\cos^2\theta+2\cos\theta)+(-\cos\theta-1)=0$$
$$2\cos\theta(\cos\theta+1)-(\cos\theta+1)=0$$
$$(\cos\theta+1)(2\cos\theta-1)=0$$
$$\cos\theta=-1\hspace{1cm}\text{or}\hspace{1cm}\cos\theta=\frac{1}{2}$$
- For $\cos\theta=-1$:
As $\theta\in[0,2\pi]$, there is only 1 point where $\cos\theta=-1$, which is $\theta=\pi$.
- For $\cos\theta=\frac{1}{2}$:
As $\theta\in[0,2\pi]$, there are 2 points where $\cos\theta=\frac{1}{2}$, which are $\theta=\frac{\pi}{3}$ and $\theta=\frac{5\pi}{3}$.
Therefore, overall, the answer to this exercise is
$$\theta=\Big[\frac{\pi}{3},\pi,\frac{5\pi}{3}\Big]$$
