Answer
The solution set is $$\{30^\circ+360^\circ n,150^\circ+360^\circ n,270^\circ+360^\circ n,n\in Z\}$$
Work Step by Step
$$2\sin\theta=2\cos2\theta$$
Here we have two different trigonometric functions, sine and cosine. It would be better if we could rewrite the equation in only one trigonometric function. Luckily, there is a way to do so.
- Recall the identity: $\cos2\theta=1-2\sin^2\theta$
which means $$2\sin\theta=2(1-2\sin^2\theta)$$
$$2\sin\theta=2-4\sin^2\theta$$
$$4\sin^2\theta+2\sin\theta-2=0$$
$$2\sin^2\theta+\sin\theta-1=0$$
$$(2\sin^2\theta+2\sin\theta)+(-\sin\theta-1)=0$$
$$2\sin\theta(\sin\theta+1)-(\sin\theta+1)=0$$
$$(\sin\theta+1)(2\sin\theta-1)=0$$
$$\sin\theta=-1\hspace{1cm}\text{or}\hspace{1cm}\sin\theta=\frac{1}{2}$$
1) First, we solve the equation over the interval $[0^\circ,360^\circ)$
- First, $\sin\theta=-1$, over the interval $[0^\circ, 360^\circ)$, there is one value of $\theta$ where $\sin\theta=-1$, which are $\theta=\{270^\circ\}$
- For $\sin\theta=\frac{1}{2}$, over the interval $[0^\circ, 360^\circ)$, there are 2 values of $\theta$ where $\sin\theta=\frac{1}{2}$, which are $\theta=\{30^\circ,150'^\circ\}$
Therefore, overall, $$\theta=\{30^\circ,150^\circ,270^\circ\}$$
2) Solve the equation for all solutions
Sine function has period $360^\circ$, so we would add $360^\circ$ to all solutions found in part 1) for $\theta$.
$$\theta=\{30^\circ+360^\circ n,150^\circ+360^\circ n,270^\circ+360^\circ n,n\in Z\}$$
This is the solution set of the given equation.
