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March 29, 2017

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Please help!!!!!!!!!!! Find all solutions to the equation in the interval [0,2π).
8. cos2x=cosx
10. 2cos^2x+cosx=cos2x
Solve algebraically for exact solutions in the interval [0,2π). Use your grapher only to support your algebraic work.
24. cos2x+sinx=0

  • Precalculus - ,

    Have you tried the double angle formulas?

  • Precalculus - ,

    Yes. It is just too hard and it would be nice if you would just explain it to me or show me an example. Please.

  • Precalculus - ,

    Hard? Why would anyone want to do something easy?

    10.2cos^2x+cosx=cos2x
    2cos^2x+cosx=cosx= cos(x)cos(x) - sin(x)sin(x) = cosĀ²(x) - sinĀ²(x)
    2 cos^2 x+cosx=cos^2x-(1-cos^2x)
    2 cos^2 x+cosx=2cos^2x-1
    cosx=-1
    x=PI

  • Precalculus - ,

    the only way to make these easy is to do dozens of them, not just the assignments. The double angle formulas are critical to know.

    24.
    cos 2x + sin x=0
    1 - 2sin^2 x + sin x = 0
    rearrange things a bit to the usual order:

    2sin^2 x - sin x - 1 = 0
    (2sin x + 1)(sin x - 1) = 0

    so, either sin x = -1/2 or sin x = 1

    sin x = -1/2: x = 7pi/6 or 11pi/6
    sin x = 1: x = pi/2

  • Precalculus - ,

    Thanks for your help you guys.

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