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22 August, 07:20

Use a graphing utility to solve the equation on the interval 0°< x < 360°. Express the solution (s) rounded to one decimal place.

cos^2 x + cos x - 1 = 0

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  1. 22 August, 07:45
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    cos^2 x + cos x - 1 = 0, and if X=cosx, x=arccosX

    let's find X

    X²+X-1=0, delta=1-4 (-1) = 5, X = - 1-sqrt5 / 2 or X = - 1 + sqrt5 / 2, but 0°< x < 360°,

    so we must take X = - 1 + sqrt5 / 2, and x=arccosX = 51.82°
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