Answer
The solution set is $$\{11.8^\circ+n180^\circ, 78.2^\circ+n180^\circ, n\in Z\}$$
Work Step by Step
$$2-\sin2\theta=4\sin2\theta$$
$$5\sin2\theta=2$$
$$\sin2\theta=\frac{2}{5}$$
1) First, we solve the equation over the interval $[0^\circ,360^\circ)$
For $\sin2\theta=\frac{2}{5}$, we have
$$2\theta=\sin^{-1}\frac{2}{5}\approx23.6^\circ$$
Also, over the interval $[0^\circ,360^\circ)$, there is one more value of $2\theta$ where $\sin2\theta=\frac{2}{5}$, which is $2\theta=180^\circ-23.6\approx156.4^\circ$
Therefore, $$2\theta=\{23.6^\circ,156.4^\circ\}$$
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 $2\theta$.
$$2\theta=\{23.6^\circ+n360^\circ,156.4^\circ+n360^\circ, n\in Z\}$$
$$\theta=\{11.8^\circ+n180^\circ, 78.2^\circ+n180^\circ, n\in Z\}$$
