How would you solve for all solutions of $\sin(2x)=\cos(3x)$ algebraically? So my buddy and I (both HS Math teachers) have been messing around with a question about finding all solutions of a "co-function" equation like the one above. The typical HS questions asks students to find the solution in Q1 by solving $2x + 3x = 90$.
He was thinking about how to get the other solutions and we were curious to see what other people do for a problem like this. 
We've noticed that the solutions come in regularly spaced intervals from two separate starting points.  
Namely, $\frac{\pi}{10} \pm \frac{4k\pi}{10}$, AND $-\frac{\pi}{2} \pm 2k\pi$  {$k$ natural} for the above equation.
We've been able to generalize this to any such co-function equation of the form $\sin(ax)=\cos(bx)$ {$a,b$ Integers} , and are curious about whether other people have stumbled upon this as well.
We've also discovered a very interesting geometric interpretation of the periodic solutions and where they occur.
We'd love to hear how you approach this type of problem (algebraically, not graphically) and whether you've encountered these ideas as well.
 A: $\sin(2x) = \cos(3x) \Rightarrow \cos(\frac{\pi}{2}-2x) = \cos (3x) \Rightarrow \dfrac{\pi}{2} - 2x = \pm 3x + 2k\pi, k \in \mathbb{Z}$. Can you take it from here?
A: There is another way. You can use the sine and cosine addition formulas to recursively factor out the whole thing. It takes a lot of patience, but at the end you get a quadratic equation you can solve.
$\sin (2x)=\cos(3x)$
$2\sin(x)\cos(x)=\cos(2x+x)$
$2\sin(x)\cos(x)=\cos(2x)\cos(x)-\sin(2x)\sin(x)$
$2\sin(x)\cos(x)=(\cos^2(x)-\sin^2(x))\cos(x)-2\sin^2(x)\cos(x)$
Then you divide everything by $cos(x)$ giving you...
$2\sin(x)=(\cos^2(x)-\sin^2(x))-2\sin^2(x)$
Use the identity $\cos^2(x)=1-\sin^2(x)$
$2\sin(x)=1-4\sin^2(x)$
Now you have something that you can substitute for to solve a quadratic. For example say $w=\sin(x)$
Then your equation looks like $2w=1-4w^2$ which is really $4w^2+2w-1=0$ in disguise. Solve the rest on your own.
A: One way is to use the identity
$$\sin \cos^{-1} = \sqrt{1 - x^2}$$
after making some substitutions.  There's a family of similar identities for other trig functions.
A: $$ \sin  2x = \sin ( \pi/2 -3 x) $$
$$ 2 x = \pi/2 - 3 x, 2 x = \pi - \pi/2 + 3 x $$ 
$$ x = \pi/10, -\pi/2 $$ 
and all co-terminal angles.
