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)
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!