Asked by Thomas
Solve the equation. (Find all solutions of the equation in the interval [0, 2π). Enter your answers as a comma-separated list.)
4 tan(2x) − 4 cot(x) = 0
4 tan(2x) − 4 cot(x) = 0
Answers
Answered by
Reiny
tan(2x) = 2tanx/(1-tan^2 x)
let tanx = k , then
4 tan(2x) − 4 cot(x) = 0
4(2k)/(1 - k^2) - 4/k = 0
8k/(1 - k^2) = 4/k
2k(1 - k^2) = 1/k
2k^2 = 1 - k^2
3k^2 = 1
k = ±1/√3
tanx = ±1/√3
so x could be in any of the 4 quadrants,
x = 30°, 150° , 210°, 330°
or
x = π/6, 5π/6, 7π/6, 11π/6
let tanx = k , then
4 tan(2x) − 4 cot(x) = 0
4(2k)/(1 - k^2) - 4/k = 0
8k/(1 - k^2) = 4/k
2k(1 - k^2) = 1/k
2k^2 = 1 - k^2
3k^2 = 1
k = ±1/√3
tanx = ±1/√3
so x could be in any of the 4 quadrants,
x = 30°, 150° , 210°, 330°
or
x = π/6, 5π/6, 7π/6, 11π/6
Answered by
Thomas
I put that in and it said it wasn't correct. Is there anything that could be missing?
Answered by
oobleck
well, π/6 works fine, since
4(√3 - √3) = 0
4(√3 - √3) = 0
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