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Trig Equation which I do not know how to solve watch

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    If anyone could help me with this it would help, thanks.


    2(sinx^2) +2sinxcosx - 1 = 0
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    How can you rewrite 2sinxcosx to get it in terms of sin?
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    Sin(A+B)=SinACosA+CosBSinB or something like that, we havent done that yet at college.
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    (Original post by ilovedubstep)
    Sin(A+B)=SinACosA+CosBSinB or something like that, we havent done that yet at college.
    you need to use it.
    2sinxcosx = sin(2x)

    When you write 2(sinx^2), what do you mean? 2(sinx)² ( = 2sin²x )?
    or 2sin(x²).
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    One way to do it:

    s= sin x
    c= cos x

    (2s^2-1)^2 = 4s^2c^2
    (2s^2-1)^2 = 4s^2(1-s^2)

    subtitute:
    y=s^2

    (2y-1)^2 = 4y(1-y)

    4y^2-4y+1 =4y-4y^2
    8y^2 - 8y + 1 = 0

    solve for y.

    x = arcsin( sqrt(y) )
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    (Original post by ilovedubstep)
    If anyone could help me with this it would help, thanks.


    2(sinx^2) +2sinxcosx - 1 = 0
    Write this as:
    2\sin x\cos x - (1-2\sin ^2x) = 0

    Then note that sin 2x = 2\sin x \cos x and \cos 2x = \cos ^2x-\sin ^2x = 1-2\sin ^2x.
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    (Original post by Melanie-v)
    you need to use it.
    2sinxcosx = sin(2x)

    When you write 2(sinx^2), what do you mean? 2(sinx)² ( = 2sin²x )?
    or 2sin(x²).

    2sin²x

    so cosxsinx = (sin(2x))/2 ?

    This leads to another problem... How do I solve

    4sin²x + sin(2x) -2 ?


    The answer is 45 dont worry I've solved it now!
 
 
 
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Updated: March 23, 2011
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