Answer
The solution set to the given equation is $$\theta=\{66.5^\circ, 112.5^\circ, 247.5^\circ,292.5^\circ\}$$
Work Step by Step
$$\sqrt2\cos2\theta=-1$$ over interval $[0^\circ,360^\circ)$
1) Find corresponding interval for $2\theta$
The interval for $\theta$ is $[0^\circ,360^\circ)$, which can also be written as the inequality:
$$0^\circ\le\theta\lt360^\circ$$
Therefore, for $2\theta$, the inequality would be
$$0^\circ\le2\theta\lt720^\circ$$
Thus, the corresponding interval for $2\theta$ is $[0^\circ,720^\circ)$.
2) Now we examine the equation: $$\sqrt2\cos2\theta=-1$$
$$\cos2\theta=-\frac{1}{\sqrt2}=-\frac{\sqrt2}{2}$$
Over interval $[0^\circ,720^\circ)$, there are 4 values whose sine equals $-\frac{\sqrt2}{2}$, which are $\{135^\circ, 225^\circ, 495^\circ,585^\circ\}$
Therefore, $$2\theta=\{135^\circ, 225^\circ, 495^\circ,585^\circ\}$$
It follows that $$\theta=\{66.5^\circ, 112.5^\circ, 247.5^\circ,292.5^\circ\}$$
This is the solution set of the equation.
