Pre-Cal(Urgent!!)

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Find all the solutions of the equation in the interval (0,2pip)

sin(x + pi/6) - sin(x -pi/6) = 1/2

I am only stuck on the last part. I have 2sinx(sqrt3/2) = 1/2 Does the 2 cancel out with the sqrt 3/2. I am not sure what to do.

  • Pre-Cal(Urgent!!) -

    I don't see how you got 2sinx(sqrt3/2) = 1/2 .

    Here is what I would do:
    sin(x + pi/6) - sin(x -pi/6) = 1/2
    sinxcospi/6 + cosxsinpi/6 - (sinxcospi/6 - cosxsinpi/6) = 1/2
    2cosxsinpi/6 = 1/2
    cosx = 1/2, since sinpi/6 = 1/2
    so x is in the first or fourth quadrants
    In quad I, x = pi/3 , (60 degrees) or
    in quad IV, x = 2pi - pi/6 = 11pi/6 , (300 degrees)

    BTW, if 2sinx(sqrt3/2) = 1/2 were correct,
    you could NOT cancel the 2's

  • Pre-Cal(Urgent!!) -

    You got a sign wrong somewhere.

    Using the sin (a+b) and sin (a-b) formulas, the left half becomes
    sinx cos pi/6 + sin(pi/6) cosx
    - sinx cos(pi/6) + cosx sin(pi/6) = 1/2
    2 cosx sin pi/6 = 1/2
    cos x = 1/2
    x = pi/3 or 5 pi/3

  • Pre-Cal(Urgent!!) -

    I agree with Reiny up until the 11 pi/6

  • Pre-Cal(Urgent!!) -

    Yup, drwls is right,

    I should have had 2pi - pi/3,

    x = 5pi/3,

    (I had 2pi - pi/6, don't know where that came from)

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