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1 September, 16:28

Solve 2cos2x + cosx - 1 = 0 for x over the interval [0, 2pi).

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  1. 1 September, 19:39
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    2cos^2x+cosx-1 = 0

    2cos^2x + 2cosx - cosx - 1 = 0

    (2cos^2x + 2cosx) - (cosx + 1) = 0

    2cosx (cosx + 1) - 1 (cosx + 1) = 0

    (cosx + 1) (2cosx - 1) = 0

    cosx+1 = 0 or 2cosx-1 = 0

    cosx+1 = 0 gives cosx = - 1, x = PI

    2cosx-1 = 0 gives cosx = 1/2, x = PI/3 or x = 5PI/3

    answer: x = PI/3 or x = PI or x = 5PI/3
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