Math Help

Hello! Can someone please check and see if I did this right? Thanks! :)

Directions: Find the exact solutions of the equation in the interval [0,2pi]


My answer:


asked by Maggie
  1. that is not an answer. It is a restatement of the problem, using only sines.

    2sin^2x-sinx-1 == 0
    (2sinx+1)(sinx-1) = 0
    sinx = -1/2 or 1, so
    x = π/2, 4π/3, 5π/3

    posted by Steve

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