Question 826960
Solve to the nearest degree if needed, for 0≤x≤360°
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For: sin^2x=cosx
1-cos^2x=cosx
cos^2x+cosx-1=0
solve for cosx by quadratic formula:
{{{cosx = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
a=1, b=1, c=-1
ans:
cosx&#8776;-1.618 (reject,(-1<cosx<1)
cosx&#8776;0.618
x&#8776;51.83&#730;,308.17&#730;
..
For:2sin^2x+cosx&#8722;cos^2x=0 
2(1-cos^x)+cosx-cos^2x=0
2-2cos^x)+cosx-cos^2x=0
3cos^2x-cosx-2=0
factor:
(3cosx+2)(cosx-1)=0
cosx=-2/3
x&#8776;131.81&#730;,228.19&#730;
or
cosx=1
x=0