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March 28, 2017

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Find all solutions on the interval (0,2pi):

2-2cos^2=sinx+1

  • Trigonometry - ,

    2 - 2(1-sin^2x) - sinx - 1 = 0
    2sin^2x - sinx -1 = 0
    (2sinx + 1 )(sinx-1) = 0
    sinx = -1/2 or sinx = 1

    x = 210° or 330° or 90°
    or
    x = 7π/6 , 11π/6, or π/2

  • Trigonometry - ,

    jkjkkllklkgjsin cos

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