Answer
The solution set is $$\{30^\circ+180^\circ n,90^\circ+180^\circ n,150^\circ+180^\circ n,n\in Z\}$$
Work Step by Step
$$2\cos^22\theta=1-\cos2\theta$$
$$2\cos^22\theta+\cos2\theta-1=0$$
$$(2\cos^22\theta+2\cos2\theta)+(-\cos2\theta-1)=0$$
$$2\cos2\theta(\cos2\theta+1)-(\cos2\theta+1)=0$$
$$(\cos2\theta+1)(2\cos2\theta-1)=0$$
$$\cos2\theta=-1\hspace{1cm}\text{or}\hspace{1cm}\cos2\theta=\frac{1}{2}$$
1) First, we solve the equation over the interval $[0^\circ,360^\circ)$
- For $\cos2\theta=-1$, over the interval $[0^\circ, 360^\circ)$, there is only one value of $\theta$ where $\cos2\theta=-1$, which is $2\theta=\{180^\circ\}$.
- For $\cos2\theta=\frac{1}{2}$, over the interval $[0^\circ, 360^\circ)$, there are two values of $\theta$ where $\cos2\theta=\frac{1}{2}$, which are $2\theta=\{60^\circ,300^\circ\}$.
Therefore, $$2\theta=\{60^\circ,180^\circ,300^\circ\}$$
2) Solve the equation for all solutions
Cosine function has period $360^\circ$, so we would add $360^\circ$ to all solutions found in part 1) for $2\theta$.
$$2\theta=\{60^\circ+360^\circ n,180^\circ+360^\circ n,300^\circ+360^\circ n,n\in Z\}$$
Finally, we find the solutions for $\theta$ by dividing all the solutions for $2\theta$ by $2$, which is also the solution set:
$$\theta=\{30^\circ+180^\circ n,90^\circ+180^\circ n,150^\circ+180^\circ n,n\in Z\}$$
