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math (trig)

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Im having problems solving this equation:

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

I keep getting an imaginary number..:/

  • math (trig) -

    2 sin x -2 = cos x = sqrt(1 - sin^2 x)
    4 sin^2 x - 8 sin x + 4 = 1 - sin^2 x
    5 sin^2 x - 8 sin x + 3 = 0
    Let sin x = y
    5y^2 -8y +3 = 0
    (5y-3)(y-1) = 0
    sin x = 1 or 3/5
    x = pi/2 (90 degrees) is one solution. There are others as well

  • math (trig) -

    im a bit confused as to where the 8sin(x) came from

  • math (trig) -

    It somes from squaring 2sin x - 2.
    [2(sin x -1)]^2 = 4(sinx-1)^2
    = 4 (sin^2 - 2x +1)= ...

  • math (trig) -

    ah, thank you, i don't know why i was squaring it differently (wrong way)

  • math (trig) -

    Jess, don't forget that after you square an equation all answers you obtain must be verified.

    for example, the second part to drwls solution was
    sinx = 3/5
    which produces answers of 36.87º and 143.13º

    when these are checked in the original equation, x= 143.13 works but x = 36.87 does not

    so x = 90º or x = 143.13º

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