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,...)
