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Math Help - Trigonemetric Function - Solving

  1. #1
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    Trigonemetric Function - Solving

    Hello,

    Can someone please help me solve these 2 problems?

    1. sqrt(3)sin2x=cos2x
    2. cos(x/2)-1=0

    Thanks!
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  2. #2
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    Q1 Rewrite so you get sin2x/cos2x = 1/sqrt(3).
    Let 2x = A and solve for A
    Then put 2x back in and solve for x
    Don't forget to consider the domain.

    Q2 Let x/2 =A, write as cosA = 1, solve for A, sub in x/2 and solve for x. Again, watch the domain.
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  3. #3
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    Quote Originally Posted by Debsta View Post
    Q1 Rewrite so you get sin2x/cos2x = 1/sqrt(3).
    Let 2x = A and solve for A
    Then put 2x back in and solve for x
    Don't forget to consider the domain.

    Q2 Let x/2 =A, write as cosA = 1, solve for A, sub in x/2 and solve for x. Again, watch the domain.
    Thanks! can you show me the algebraic step for Q2?

    The answer is 4kpi...

    mine looks like this:

    cos(x/2)-1=0
    cos(x/2)=1 which is pi +kpi
    cos(x/2)= (pi + kpi)2
    cosx=2pi+2kpi
    is that right so far?
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  4. #4
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    Quote Originally Posted by l flipboi l View Post
    Thanks! can you show me the algebraic step for Q2?

    The answer is 4kpi...

    mine looks like this:

    cos(x/2)-1=0
    cos(x/2)=1 which is pi +kpi No, This means that x/2 = 2k pi, because
    cos(even multiples of pi)=1 and cos(oddmultiples of pi) = -1. The 2 in the 2k pi means even multiples of pi
    So x/2 = 2k pi
    So x = 4k pi


    cos(x/2)= (pi + kpi)2 Once you've brought in your pi's you don't have the cos there anymore.

    cosx=2pi+2kpi and you can't say that if cos(x/2) = n then cos x = 2n You just can't take the 2 out like that.
    is that right so far?
    Hope that helps
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  5. #5
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    Thanks!
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