Exact Solution

QwertyKnight

New member
Joined
May 3, 2015
Messages
7
Hi guys,
I solved tan(2x)=cot(x), and i only got 2 solutions(+-(pi/6)+n*pi), but there's a third solution(pi/2 + k*pi). I already checked lot of websites, but on those websites there were only 2 solutions. How would i calculate that third solution? I would appreciate the help!
 
Hi guys,
I solved tan(2x)=cot(x), and i only got 2 solutions(+-(pi/6)+n*pi), but there's a third solution(pi/2 + k*pi). I already checked lot of websites, but on those websites there were only 2 solutions. How would i calculate that third solution? I would appreciate the help!

\(\displaystyle \displaystyle{tan(2x) \ = cot(x)}\)

\(\displaystyle \displaystyle{tan(2x) \ - \ cot(x) \ = \ 0}\)

\(\displaystyle \displaystyle{\frac{2tan(x)}{1-tan^2(x)} \ - \ \frac{1}{tan(x)} \ = \ 0}\)

\(\displaystyle \displaystyle{\frac{3tan^2(x) \ - \ 1}{tan(x) * [1-tan^2(x)]} \ = \ 0}\)

The condition above will be true when tan(x) = ∞ → x = π/2 + k*π (in addition to the condition when the numerator = 0).
 
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