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Math Help - number of solutions within 360 degrees

  1. #1
    Senior Member furor celtica's Avatar
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    number of solutions within 360 degrees

    Solve the following equation of A, giving solutions in the 0<A<360 inclusive (i don't know how to make the 'or equal to' sign):
    sin2A - (sqrt3)(cos2A) =0
    So I did it like this
    Taking B=2A
    sinB - (sqrt3)(sqrt(1-(sinB)^2)) = 0 (because cosB = sqrt(1-(sinB)^2)
    sinB=sqrt(3-3(sinB)^2)
    (sinB)^2=3-3(sinB)^2
    4(sinB)^2=3
    sinB=(plus or minus)sqrt3/2
    from this I get the results A=30, 60, 120, 150, 210, 240, 300, 330
    However, only 30, 120, 210 and 300 are correct.
    Why is this?
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  2. #2
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    mr fantastic's Avatar
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    Quote Originally Posted by furor celtica View Post
    Solve the following equation of A, giving solutions in the 0<A<360 inclusive (i don't know how to make the 'or equal to' sign):
    sin2A - (sqrt3)(cos2A) =0
    So I did it like this
    Taking B=2A
    sinB - (sqrt3)(sqrt(1-(sinB)^2)) = 0 (because cosB = sqrt(1-(sinB)^2)
    sinB=sqrt(3-3(sinB)^2)
    (sinB)^2=3-3(sinB)^2
    4(sinB)^2=3
    sinB=(plus or minus)sqrt3/2
    from this I get the results A=30, 60, 120, 150, 210, 240, 300, 330
    However, only 30, 120, 210 and 300 are correct.
    Why is this?
    Extraneous solutions introduced by squaring and then square rooting. But why would you do it this way?

    Much simpler to re-arrange:

    \sin(2A) = \sqrt{3} \cos(2A) \Rightarrow \tan(2A) = \sqrt{3} which is simple to solve.
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  3. #3
    Senior Member furor celtica's Avatar
    Joined
    May 2009
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    yes i thought it was that but wasn't sure how it happened.
    good point with the simpler solution. not very clever of me.
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