2cos^2x+cosx-1=0
find the degree solutions

2cos²x+cos x-1=0

cos x=t

2t²+t-1=0

t=[-1⁺₋√(-1+8)]/2=(-1⁺₋3)/4
We have two possible set solutions:
First set solutions.
t₁=(-1-3)/2=-4/4=-1   
cos x=-1  ⇒x=cos⁻¹ (-1)=π +2kπ    or     180º+360ºk    (k=(...-2,-1,0,1,2...)

Second set solutions:
t₂=(-1+3)/4=2/4=1/2

cos x=1/2  ⇒ x=cos⁻¹ 1/2=π/3+2kπ   U   5π/3+ 2kπ   or
60º+360ºK  U 300º+360ºK        (k=...-2,-1,0,1,2,...)

solutions: first set solutions U second set solutions:

Answer in radians : π +2kπ U π/3+2kπ U 5π/3+ 2kπ   (k=...-1,0,1,...)
Answer is degrees: 180º+360ºk U 60º+360ºK  U 300º+360ºK        (k=...-2,-1,0,1,2,...)

Answer is a not really sure tho

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