Question

How many solutions does cos2x=sinx+cosx have?

Are you supposed to use the double angle formula?

Answer 1

\[\cos2x=\cos(x+x)=\cos x cosx-sinxsinx=\cos^2x-\sin^2x\]
So \[\cos^2x-cosx=sinx+\sin^2x\]
No, try again:
\[\cos^2x-\sin^2x=(cosx+sinx)(cosx-sinx)\]\[=cosx+sinx\]
so cosx - sinx = 1
But this is always true, so infinitely many values of x between zero degrees and 90 degrees.
I think!

Answer 2

There are going to be infinite solutions because the left side alternates between 1 and -1... with the right alternating between ~1.5 and -1.5...when they cross is a bit harder to pin down--but there will be infinite solutions.