trig
posted by Anonymous .
determine the solutions for tanxcos^2xtanx=0 in the interval XE[2pi,2pi]
Ok, please describe the steps to me that you used to solve this! thank you

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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