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put this solution on YOUR website!Find all values of x in interval [0, 2(pi)] that satisfy
sin2(x)=cos(x).
1-cos^2x=cosx
cos^2x+cosx-1=0
Solve for cosx by quadratic formula:
a=1, b=1, c=-1
cosx=[-1±√(1-4*1*-1)]/2*1
cosx=(-1±√5)/2
cosx=(-1±2.236)/2
cosx=-1.618 (reject, cosx≥-1)
cosx=.618
x≈.905 and 5.378 radians in quadrants I and IV where cos>0