Question 797886
Solve algebraically cos^2 x = sin x to find solutions in the domain 0<= x < 2 pi
cos^2 x = sin x
1-sin^2 x=sin x
sin^2 x+sin x-1=0
solve for sin x by quadratic formula:
{{{sin x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
a=1, b=1, c=-1
sin x=-1.618 (reject, -1 < sin x < 1)
or
sin x=0.618
x&#8776;0.6661