Solve the equation algebraically for x wohere 0 </ x< 2 pi.

Question

sin2 x = cos x

Answer 1

sin^2 x=cosx ?

1-cos^2 x=cosx
0=cos^2 x + cosx -1

cosx=(-1+-sqrt(1+4))/2=-1/2 +-1/2 sqrt 5

x= arccos(-.5-.5sqrt5) or arccos(-.5+.5sqrt5) one of these does not work

then x=51.8deg

check my math