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Math Help - trig: find x's

  1. #1
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    trig: find x's

    Solve for x E [0,2pi]: sin(3x) - sin(x) = cos(2x)
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Mr_Green View Post
    Solve for x E [0,2pi]: sin(3x) - sin(x) = cos(2x)
    let's expand this

    sin(3x) - sin(x) = cos(2x)
    => sin(2x)cos(x) + sin(x)cos(2x) - sin(x) = cos^2(x) - sin^2(x)
    => [2sin(x)cos(x)]cos(x) + sin(x)[cos^2(x) - sin^2(x)] - sin(x) = cos^2(x) - sin^2(x)
    => 2sin(x)cos^2(x) + sin(x)[cos^2(x) - sin^2(x)] - sin(x) = cos^2(x) - sin^2(x)

    we have a lone sin(x), so let's change everything to sin(x)

    => 2sin(x)(1 - sin^2(x)) + sin(x)[1 - sin^2(x) - sin^2(x)] - sin(x) = 1 - sin^2(x) - sin^2(x)
    => 2sin(x) - 2sin^3(x) + sin(x) - 2sin^3(x) - sin(x) = 1 - 2sin^2(x)
    => 2sin(x) - 2sin^3(x) + sin(x) - 2sin^3(x) - sin(x) - 1 + 2sin^2(x) = 0
    now this is ugly, let's try and simplify this a bit

    => -4sin^3(x) + 2sin^2(x) + 2sin(x) - 1 = 0
    this is a cubic in sin(x), to cut down on typing for the mean time, let's write sin(x) as y, we get:

    -4y^3 + 2y^2 + 2y - 1 = 0 ........ah, that looks a little better
    => -2y^2(2y - 1) + (2y - 1) = 0
    => (2y - 1)(1 - 2y^2) = 0
    => y = 1/2 or y = +/- sqrt(1/2)

    but, y = sin(x)
    => sin(x) = 1/2 or sin(x) = +/- sqrt(1/2) = +/- 1/sqrt(2) = +/- sqrt(2)/2
    => x = pi/6, 5pi/6 or x = pi/4, 3pi/4, 5pi/4, 7pi/4

    so there are 6 values of x for which this happens
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    hmm, it seems i've made an error somewhere. only two of my values are actual solutions (i graphed the functions to see)

    there are 6 values though. pi/6 and 5pi/6 work
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  4. #4
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    Hello, Mr_Green!

    Solve for x ε [0,2π]: .sin(3x) - sin(x) .= .cos(2x)
    We need a sum-to-product identity:
    . . sin(A) - sin(B) .= .2·cos(½(A + B)]·sin[½(A - B)]

    The left side is:
    . . sin(3x) - sin(x) .= .2·cos[½(3x + x)]·sin[½(3x-x)] .= .2·cos(2x)·sin(x)


    The equation becomes: .2·cos(2x)·sin(x) .= .cos(2x)

    . . . . . . . . . . .2·cos(2x)·sin(x) - cos(2x) .= .0

    Factor: .cos(2x)·[2·sin(x) - 1] .= .0


    We have two equations to solve:

    cos(2x) = 0 . . 2x = π/2, 3π/2, 5π/2, 7π/2 . . x = π/4, 3π/4, 5π/4, 7π/4

    2·sin(x) - 1 .= .0 . . sin(x) = ½ . . x = π/6, 5π/6

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  5. #5
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    don't worry about it. i will look through your work, and let you know if I find anything.

    thanks
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  6. #6
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    Jhevon I checked your work and everything is in order. i also graphed them and they were all correct.
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