Answer
The solution set to this problem is $$\{90^\circ,270^\circ\}$$
Work Step by Step
$$\sin^2\theta\cos\theta=\cos\theta$$ over interval $[0^\circ,360^\circ)$
1) Solve the equation:
$$\sin^2\theta\cos\theta=\cos\theta$$
Again, 2 different trigonometric function types would not help, so we need to find a way to eliminate either sine or cosine function, or separate them into different equations.
$$\cos\theta-1\sin^2\theta\cos\theta=0$$
$$\cos\theta(1-\sin^2\theta)=0$$
Recall the identity: $1-\sin^2\theta=\cos^2\theta$
$$\cos^3\theta=0$$
$$\cos\theta=0$$
2) Apply the inverse function:
Over the interval $[0^\circ,360^\circ)$, examining the unit circle, it is found that there are 2 values of $\theta$ where $\cos\theta=0$, which are $\{90^\circ,270^\circ\}$
Therefore, $$\theta\in\{90^\circ,270^\circ\}$$
In other words, the solution set to this problem is $$\{90^\circ,270^\circ\}$$
