trig

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determine the solutions for tanxcos^2x-tanx=0 in the interval XE[-2pi,2pi]

Ok, please describe the steps to me that you used to solve this! thank you

  • trig -

    tanx(cos^2 x - 1) = 0
    tanx = 0 or cosx = ± 1

    looking at tangents curves from -2pi to +2pi
    x = -2pi , -pi, 0, pi, 2pi

    looking at a cos curve where it is ±1
    is -2pi, -pi 0, pi, and 2pi

    so both parts had the same solutions,
    x = 0, ±pi, ±2pi

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