Answer
The solution set to this problem is $$\{45^\circ, 135^\circ, 225^\circ, 315^\circ\}$$
Work Step by Step
$$\cos^2\theta-\sin^2\theta=0$$ over interval $[0^\circ,360^\circ)$
1) Solve the equation:
$$\cos^2\theta-\sin^2\theta=0$$
- Recall the identity: $\sin^2\theta=1-\cos^2\theta$
$$\cos^2\theta-(1-\cos^2\theta)=0$$
$$\cos^2\theta-1+\cos^2\theta=0$$
$$2\cos^2\theta-1=0$$
$$\cos^2\theta=\frac{1}{2}$$
$$\cos\theta=\pm\frac{1}{\sqrt2}=\pm\frac{\sqrt2}{2}$$
2) Apply the inverse function:
- For $\cos\theta=\frac{\sqrt2}{2}$: Over the interval $[0^\circ,360^\circ)$, $\cos\theta\gt0$ means the value of $\theta$ lies either in quadrant I or IV.
In fact, the angles of $45^\circ$ in quadrant I and $315^\circ$ in quadrant IV are where cosine equals $\frac{\sqrt2}{2}$.
- For $\cos\theta=-\frac{\sqrt2}{2}$: Over the interval $[0^\circ,360^\circ)$, $\cos\theta\lt0$ means the value of $\theta$ lies either in quadrant II or III.
In fact, the angles of $135^\circ$ in quadrant II and $225^\circ$ in quadrant III are where cosine equals $-\frac{\sqrt2}{2}$.
Therefore, $$\theta\in\{45^\circ, 135^\circ, 225^\circ, 315^\circ\}$$
In other words, the solution set to this problem is $$\{45^\circ, 135^\circ, 225^\circ, 315^\circ\}$$
