Question:
Find all angles $x$, $0^\circ \le x < 180^\circ,$ such that
\[\sin 6x + \cos 4x = 0.\]Enter all the solutions, separated by commas.

Answer:
We can write
\[\sin 6x + \cos 4x = \sin 6x + \sin (90^\circ - 4x).\]Then from the sum-to-product formula,
\begin{align*}
\sin 6x + \sin (90^\circ - 4x) &= 2 \sin \left( \frac{6x + 90^\circ - 4x}{2} \right) \cos \left( \frac{6x - (90^\circ - 4x)}{2} \right) \\
&= 2 \sin (x + 45^\circ) \cos (5x - 45^\circ).
\end{align*}Thus, $\sin (x + 45^\circ) = 0$ or $\cos (5x - 45^\circ) = 0.$

If $\sin (x + 45^\circ) = 0,$ then $x = 135^\circ.$

If $\cos (5x - 45^\circ) = 0,$ then $5x - 45^\circ$ must be $90^\circ,$ $270^\circ,$ $450^\circ,$ $630^\circ,$ or $810^\circ.$  These lead to the solutions $\boxed{27^\circ, 63^\circ, 99^\circ, 135^\circ, 171^\circ}.$