solve $\sin(3x - 4) = \cos(7x)$ I am attempting to solve the equation $$\sin(3x + 4) = \cos(7x),$$ with all numbers in degrees.

My process is as such:

$\cos(7x) = \sin(90 - 7x)$

$\sin(3x + 4) = \sin(90 - 7x)$

$3x + 4 = 90 - 7x + 360n$; (where n is an integer and $360n$ is added due to cycling)

$x = \frac{94}{10} + 36n$

However, when I graph, I see that there is another answer that I have not solved for, which is $\frac{133}{2} + 90n$. How can I achieve this answer?
Draw a graph of $\sin x$. 

You will quickly see that $\sin a = \sin b$ does not imply $a \equiv b \pmod{360}$ - just see $\sin 60 = \sin 120$. 

The rest is left to the reader.
$$\sin(3x+4)=\sin3x\cos4+\cos3x\sin4$$ and \begin{align}\cos(7x)+\cos90&amp;=2\cos\frac{(90+7x)}{2}.\cos\frac{(90-7x)}{2}\\
&amp;=\cos^2(45)-\sin^2(7x/2)\\
&amp;=\frac{1}{4}-\sin^2(7x/2)\end{align} now you have two sin equations can you the graphs from here if you think cos and sin are bit difficult.