Originally Posted by

**guest12345** yea but that's not what I was meaning, I was meaning if there's an equation with sin(4x) = cos(5x) (as posted in my new thread) or something of that nature then how do i solve it not for the equation in the post above, was just meaning that sort of equation in general, sorry if I was unclear.

Then you can simply use $\displaystyle \cos x = \sin \left(x+\frac{\pi}{2}\right)

$ :

$\displaystyle

\sin(4x)=\cos(5x) \Longleftrightarrow \sin(4x)=\sin\left(5x+\frac{\pi}{2} \right)\Longleftrightarrow x\equiv\ldots$