Precalculus(NEED HELP ASAP PLEASE!!)

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Solve:
sin(theta)cos(theta)-sin(theta)−cos(theta)+1 = 0, 0 < or equal to theta < 2pi

Solution Set: ____________ ?

  • Precalculus(NEED HELP ASAP PLEASE!!) -

    sin(theta)cos(theta)-sin(theta)−cos(theta)+1 = 0

    sinT(sqrt(1-sin^2T)) -sinT -sqrt(1-sin^2T)+1 = 0
    let x = sinT
    x sqrt(1-x^2) -x -sqrt(1-x^2) +1 = 0

    (x-1) sqrt(1-x^2) = x-1

    sqrt (1-x^2) = (x-1)/(x-1) =1

    1-x^2 = 1 or -1

    sin^2 T = 0
    or
    sin^2 T = 2 which is impossible because sin T is between -1 and +1

  • changed last lines -

    sqrt (1-x^2) = (x-1)/(x-1) =1

    1-x^2 = 1

    sin^2 T = 0

    T = 0 or 180 degrees, 0 or pi

    check 0 and pi in the original

  • check -

    check T= 0 solution
    sin(theta)cos(theta)-sin(theta)−cos(theta)+1

    0 (1) - 0 - 1 + 1 = 0 check so zero works

    check T = pi solution

    0 -0 - (-1) + 1 = 0

    2 = 0 NO, so pi is not a solution.

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