calc

2cos2x=-sinx
solve for x over the interval [-pi,+pi]

x= 0.6349radians
x=2.507radians
x= 1.003 radians
x= 2.138 radians

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  1. 2cos2x=-sinx
    2cos 2x + sinx = 0
    2( 1 - 2sin^2 x) + sinx = 0
    4sin^2 x - sinx - 2 = 0
    sinx = (1 ± √1 - 4(4)(-2)) /8
    = (1 ± √33)/8
    = appr .84307.... or appr -.59307...

    If sinx = .84307
    x is in quads I or II, and x = appr 1.002967 or appr. 2.1386

    if sinx = -.59307..
    then x is in quads III or IV
    x = 3.776459.. which is beyond the given domain
    or
    x = 5.6483, beyond the given domain
    how about 5.6483 - 2π= -.634867 ?? , that fits the domain. , you did not have that.
    how about 3.776459 - 2π = -2.5067 ? YUP, that works also.

    so x = appr. 1.002967, 2.1386, -.634867, and -2.5067

    as confirmed by:
    http://www.wolframalpha.com/input/?i=y+%3D+2cos+(2x)+%2B+sinx

    showing 2 positve answers and 2 negative answers in
    [-π,+π]

    (I also checked these 4 answers with a calculator. You have shown 4 answers that are all positive. The graph contradicts that)

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  2. 2cos2x=-sinx
    2cos 2x + sinx = 0
    2( 1 - 2sin^2 x) + sinx = 0
    4sin^2 x - sinx - 2 = 0

    it should put
    4sin^2 x - sinx + 2 = 0

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  3. so if c=2
    then the values have to be x= 0.6349radians
    x=2.507radians
    x= 1.003 radians
    x= 2.138 radians

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  4. If calc mean calculus then:

    2 cos ( 2x ) = - sinx

    2 cos ( 2x ) + sinx = 0

    f(x) = sin x + 2 cos ( 2x )

    f´(x) = cos x - 4 cos ( 2x )

    x(n+1) = xn - [ sin xn + 2 cos ( 2xn ) ] / [ cos xn - 4 cos ( 2 xn ) ]

    This is very, very difficult for calculate.

    If you in wolframalpha. c o m type:

    solve 2 cos( 2x ) = - sin ( x ) for x = - pi to pi

    The solutions are:

    -0.634867

    -2.50673

    1.00297

    2.13863

    In your list only x= 1.003 radians and x= 2.138 radians are correct solutions.

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  5. Nope, my equation is correct, I should not have skipped that one step:

    2cos2x=-sinx
    2cos2x + sinx = 0
    2(1 - 2sin^2 x) + sinx = 0
    2 - 4sin^2 x + sinx = 0
    times -1
    -2 + 4sin^2 x - sinx = 0
    4sin^2 x - sinx - 2 = 0

    checking one of your answers in the original equation:
    let x = 2.5067
    Left Side = 2cos(5.0134
    = 2(.29648...)
    = appr .59297...
    Right Side = -sin(2.5067)
    = appr. -.593
    ≠ left side, so it does not work,
    Try mine, it works, as shown in the graph.

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  6. Final word:

    http://www.wolframalpha.com/input/?i=solve+2cos(2x)%3D-sinx

    look at the graph, let your mouse hover over the red intersection points.

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  7. My correction:

    x = 2.138 radians isn't correct solutions becouse:

    2.13863 ≈ 2.139

    Only x= 1.003 radians is correct solutions becouse:

    1.00297 ≈ 1.003

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  8. 2 - 4sin^2 x + sinx = 0
    -4sin^2x + sin x + 2 = 0
    a= -4
    b= 1
    c= 2

    why are you multiplying by -1?

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  9. Because it is customary, and it often makes it easier, if the coefficient of the square term is positive.

    So I switched all the signs, you must have known that this would not change the solution to the equation.

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  10. lol i see my mistake
    i divided by 8 and not -8

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