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, ...)