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    Shiela123's Avatar
    Shiela123 Posts: 1, Reputation: 1
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    #1

    Dec 28, 2010, 07:19 PM
    Algebra II/Trig
    Find, to the nearest degree, all values of x in the interval 0° ≤ x ≤ 360° that satisfy the equation 3 cos 2x + cos x + 2 = 0.
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #2

    Dec 29, 2010, 07:38 AM

    Suggestion: recall that cos2x can be rewriiten as one of the following:

    cos^2x - sin^2x, or
    1-2sin^2x, or
    2cos^2x - 1

    Note that if you substitute the 3rd version into the original equation you can recast the original equation as a quadratic, meaning in the form Acos^2x + Bcos^x + C = 0. Then you can factor this to find values of cosx, and from that determine values for x. You should find that there are 2 values of cos(x), and hence 4 possible solutions for x.

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