Question 944123
determine general solution in radians for sin(2x)sinx=cos^2x
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sin(2x)sinx=cos^2x
2sinxcosxsinx=cos^2x
2sin^2x=cosx
2(1-cos^2x)=cosx
2-2cos^2x=cosx
2cos^2x+cosx-2=0
solve for cosx by quadratic formula:
{{{cosx = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
a=2, b=1, c=-2
ans: 
cosx=-1.28 (reject)
or
cosx=0.78
x=0.68 radians