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10 March, 00:21

2cos^2x+cosx-1=0

Find the degree solutions

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  1. 10 March, 00:36
    0
    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, ...)
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