equations trig

2cos^2x-3sinx=0

asked by LiNnda
  1. 2(1 - sin^2x) - 3sinx = 0
    -2sin^2x - 3sinx + 2 = 0
    2sin^2x + 3sinx - 2 = 0
    (2sinx -1)(sinx + 2) = 0
    sinx = 1/2 or sinx = -2
    x = 30° or 150° or (sinx = -2 has no solution)

    x = 30 or 150 ° which is also π/6 or 5π/6 radians

    posted by Reiny

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