TITLE: Solve $\cos{(7x)}+\sin{(3x)}=0$
QUESTION [7 upvotes]: Solve $\cos(7x)+\sin(3x)=0$
I'm stuck. Help me, please. 
I did $\cos(7x)=\cos(4x+3x)=\cos(4x)\cos(3x)-\sin(4x)\sin(3x)$
So the original equation becomes
$\cos(4x)\cos(3x)-\sin(4x)\sin(3x)+\sin(3x)=0$
But that became long and ugly. What can I do now?

REPLY [3 votes]: Well, since no one else has, here's the prosthaphaeresis solution: we have
$$ \cos{7x}+\cos{(\tfrac{1}{2}\pi-3x)} = 0, $$
using $\cos{x}=\sin{(\frac{1}{2}\pi-x)}$. We then have the prosthaphaeresis formula
$$ \cos{A}+\cos{B} = 2\cos{\frac{A+B}{2}}\cos{\frac{A-B}{2}}, $$
which we use to write the equation in the form
$$ 0 = \cos{\left( \frac{7x-3x+\pi/2}{2} \right)}\cos{\left( \frac{7x+3x-\pi/2}{2} \right)} = \cos{\left( 2x+\frac{\pi}{4} \right)}\cos{\left( 5x-\frac{\pi}{4} \right)} $$
Therefore the equation is satisfied precisely when one of these factors is zero. $\cos{A}$ has roots when $A=(n+1/2)\pi$ for integer $n$, and I'm sure you can take it from here.