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Math Help - Solving Trigonometric equations

  1. #1
    Member
    Joined
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    Solving Trigonometric equations

    Question

    Find the angles too?
     2sin^2t-cost-1=0 if 0 \leq \pi \leq 2\pi

    Attempt

     cos^2t-cost-1=0

     t=2 ;t=-1

     arccos(2)=math error ; arccos(-1)=180


    Is this correct?



    Thank you
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  2. #2
    MHF Contributor
    Joined
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    Talking

    Quote Originally Posted by mj.alawami View Post
     2sin^2t-cost-1=0 if 0 \leq \pi \leq 2\pi
    How did "2sin^2(t)" become "cos^2(t)"...?

    Instead, try using the Pythagorean Identity in the form sin^2(t) = 1 - cos^2(t), so you get:

    . . . . . 2(1\, -\, \cos^2(t))\, -\, \cos(t)\, -\, 1\, =\, 0

    . . . . . 2\, -\, 2\cos^2(t)\, -\, \cos(t)\, -\, 1\, =\, 0

    . . . . . 0\, =\, 2\cos^2(t)\, +\, \cos(t)\, -\, 1

    . . . . . 0\, =\, (2\cos(t)\, -\, 1)(\cos(t)\, +\, 1)

    . . . . . \cos(t)\, =\, \frac{1}{2}\, \mbox{ or }\, \cos(t)\, =\, -1

    ...and so forth.
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