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Problem with solutions to trig equations watch

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    Hey guys just a slight issue with the question,
    sqrt3sinx=cosx
    The obvious way to solve is to divide by cos x and get
    Sqrt3tanx=1
    And solve for x

    However, when solving it I did this,
    Sqrt3sinx-cosx=0

    Sinx=cosxtanx

    Sqrt3cosxtanx - cosx = 0

    Factoring out cos x

    Cosx( sqrt3tanx - 1) = 0

    So you'll still get the same solutions from the tan part as before, but new ones are introduced.

    Between 0 and 360 degrees you have solutions for cosx at 90 and 270, neither of which work in the original equation.

    Why is this?

    Thanks, sorry for the long post
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    Could you explain the equation a bit more? Do you mean Square root of (3sinx) = cosx
    or cube root(sinx) = cosx or something else entirely?
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    (Original post by Tskadem)
    Hey guys just a slight issue with the question,
    sqrt3sinx=cosx
    The obvious way to solve is to divide by cos x and get
    Sqrt3tanx=1
    And solve for x

    However, when solving it I did this,
    Sqrt3sinx-cosx=0

    Sinx=cosxtanx

    Sqrt3cosxtanx - cosx = 0

    Factoring out cos x

    Cosx( sqrt3tanx - 1) = 0

    So you'll still get the same solutions from the tan part as before, but new ones are introduced.

    Between 0 and 360 degrees you have solutions for cosx at 90 and 270, neither of which work in the original equation.

    Why is this?

    Thanks, sorry for the long post
    Hmmm. Unfortunately this is one of those things with trig. It is easy to create phantom solutions, and you should always check your solutions.

    Notice that your equation:

    cosx(sqrt3tanx - 1) = 0 is just the equation sqrt3tanx - 1 = 0, multipied by cos x. You could multiply the equation by anything, and provided that thing can equal zero, you'll get phantom solutions.

    For example, you could have had sinx(sqrt3tanx - 1) = 0, which would have given you the solutions 0 and 180, which are obviously wrong.

    Nothing wrong with your method, and provided you check them, you'll get the correct solutions.

    Sometimes you can all but guarantee phantom solutions, eg if you square both sides of an equation. This one is not as clear, but changing one trig term into two might have been an indicating factor.

    Sorry its not a full explanation!
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    (Original post by Tskadem)
    Hey guys just a slight issue with the question,
    sqrt3sinx=cosx
    The obvious way to solve is to divide by cos x and get
    Sqrt3tanx=1
    And solve for x

    However, when solving it I did this,
    Sqrt3sinx-cosx=0

    Sinx=cosxtanx

    Sqrt3cosxtanx - cosx = 0

    Factoring out cos x

    Cosx( sqrt3tanx - 1) = 0

    So you'll still get the same solutions from the tan part as before, but new ones are introduced.

    Between 0 and 360 degrees you have solutions for cosx at 90 and 270, neither of which work in the original equation.

    Why is this?

    Thanks, sorry for the long post
    When you multiply by something you can add in extra solutins

    It should be clear that cosx cannot equal zero as cos and sin cannot be zero at the same time
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    Okay, thanks guys


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