I'm having trouble figuring out how to solve cos x = 2sinxcosx within 0 to 2pi

I know that 2sinxcosx is an identity for sin2x but that doesn't really help the situation since I want to let cosx be a variable (ie. let cosx = m and then solve for roots/angles)

I was thinking about factoring the second side of the equation but it seems like I can't get anything out of that either, I would just get

cos x = sinx (2cosx)

or

cos x = cosx (2sinx)

which doesn't really do anything

Anyways if anyone can help solve it by letting either cosx or sinx being a variable it would greatly be appreciated

The answer is supposedly 3pi/2, thanks for the help in advance!