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Math

posted by .

the original problem was:

Solve: sin(3x)-sin(x)=cos(2x)

so far i've gotten to:
sin(x)(2sin(x)cos(x)-1)=cos^2(x)-sin^2(x)

Where would I go from here?

  • Math -

    you cannot do that!
    looks like you factored
    sin(3x) - sinx = sinx(sin(2x) - 1)
    not true!!!!!!

    then sin 90 - sin 30 = sin30(sin60 - 1) Is it???, of course not

    we have a formula for sin (3x)
    = 3(cos^2 x)(sinx) - sin^3 x
    then
    sin(3x)-sin(x)=cos(2x)
    3(cos^2 x)(sinx) - sin^3 x = 1 - 2sin^2 x
    replace cos^2 x with 1 - sin^2x, expand and simplify to get
    -4sin^3 x + 2sin^2 x + 2sinx - 1 = 0
    2sin^2x(-2sinx + 1) -1(-2sinx + 1) = 0
    (-2sinx + 1)(2sin^2x - 1) = 0
    sinx = 1/2 or sinx = ± 1/√2

    x = 30, 150, 45, 135 degrees

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