# 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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