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Trigonometric Solutions

  1. Aug 12, 2017 #1
    1. The problem statement, all variables and given/known data
    The number of solutions of the equation sin2x –2cosx + 4sinx = 4 in the interval [0, 5π] is what?

    2. Relevant equations
    sin2x=2sinxcosx

    3. The attempt at a solution
    sin2x-2cosx+4sinx=4
    ⇒2sinxcosx-2cosx+4sinx=4
    ⇒sinxcosx-cosx+2sinx=2
    ⇒cosx (sinx-1)=2-2sinx
    ⇒cosx (sinx-1)=2 (1-sinx)
    ⇒cosx (sinx-1)= -2 (sinx-1)
    ⇒cosx = -2
    ∴ There is no solution
     
  2. jcsd
  3. Aug 12, 2017 #2

    Orodruin

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    Your last step assumes something that is not necessarily true.
     
  4. Aug 12, 2017 #3

    berkeman

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    WolframAlpha finds at least one solution in [0,π]...
     
  5. Aug 13, 2017 #4

    scottdave

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    Do you have the ability to plot the expression on the left side (with a calculator or spreadsheet)? Then see if it crosses 4 anywhere. That is a start to see how many solutions you are dealing with.

    Like @Orudruin indicated - you divided both sides by an expression. There is a time when you don't want to do that.
     
  6. Aug 14, 2017 #5

    ElectricRay

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    I found indeed several with my calculator but solving it algebraically

    After the step cosx (sinx-1)=2 (1-sinx) as the OP did i got lost. Im strying to solve it as well without solving it for the OP. But why you can't devide after this step, what is wrong about that?
     
  7. Aug 14, 2017 #6

    Orodruin

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    First of all, be aware that posting complete solutions before the OP has solved the problem is not allowed. See the forum rules.

    Regarding your question: In which cases is it not ok to divide by a real number ##x##?
     
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