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    • Thread Starter
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    I this looks easy, but how do I solve:

    sin2x = sinx (0< x < 360)

    I don't know how to re-aarange it :confused:



    also....
    how do I re-arrange: 8sin^3 x - 6sinx + 1 = 0, to solve it.
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    Start with the first one.

    use the double angle formula

    sin2x=2sinxcosx
    so you've got 2sinxcosx=sinx
    Then divide both sides by sinx
    2cosx=1
    cosx=1/2
    And then solve it between 0 and 360.
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    sin2x = 2sinxcosx

    therefore, 2sinxcosx = sinx

    2sinxcosx - sinx = 0

    sinx(2cosx-1) = 0

    sinx = 0,
    2cosx=1, cosx=1/2

    i think thats right...
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    Your first one is a double angle formula.
    sinx(2x)=2sinxcosx then work on from there...

    The second one, let sinx = X then you get the cubic
    8X^3 -6X +1=0, find the roots of it, which are equal to sinx. arcsin the roots then you have the solutions.
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    (Original post by Manitude)
    Start with the first one.

    use the double angle formula

    sin2x=2sinxcosx
    so you've got 2sinxcosx=sinx
    Then divide both sides by sinx
    2cosx=1
    cosx=1/2
    And then solve it between 0 and 360.
    You actually don't want to divide by sinx, you lose solutions like that.

    Just factorise it sinx(2cosx-1)= 0 and solve for sinx =0 and cosx =1/2
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    (Original post by Manitude)
    Start with the first one.

    use the double angle formula

    sin2x=2sinxcosx
    so you've got 2sinxcosx=sinx
    Then divide both sides by sinx
    2cosx=1
    cosx=1/2
    And then solve it between 0 and 360.
    You lose a solution by dividing by sinx..

    you need to make it 2sinxcosx - sinx = 0 and then factorise that to get all solutions..
    • Community Assistant
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    yeah...good point, Clarity Incognito and andrew.P...I should have noticed that.
 
 
 
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