Find the general solution of tan2x = -cot(x+pie/3)?

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Find the general solution of tan2x = -cot(x+pie/3)?

Mathematics
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u kno tan2x = 2tanx/1-tan^2x nd cotx = 1/tanx
form a quadratic nd solve........ for tanx........ then u l get the g.s
\[\tan(2x) = -\cot(90 + x -30) \implies -(-\tan(x-30)) = \tan(x - \frac{\pi}{6})\] \[\implies \tan(2x) = \tan(x - \frac{\pi}{3})\]

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Other answers:

General solution is given by: \[\tan ( \theta) = \tan(x)\] then: \[\theta = n \pi + x\]
so: \[2x = n \pi + ( x - \frac{\pi}{3} ) \implies x = n \pi + \frac{\pi}{3}\]
hey @waterineyes but my books says it as x = n pie + 5pie/6
So put n = 0 1 2 3 4 etc etc and find the corresponding value of x..
you can subtract it from \(2 \pi\)..
wat???
Wait something got wrong.. Let me check..
@w\ r uthere
Wait I am checking it..
I got it..
|dw:1343210793286:dw|
\[\tan(2x) = \cot(-x-60) \implies \cot(90 - (x + 150)) \implies \tan(x + 150)\] \[\tan(2x) = \tan(x + 150)\]
150 is in degrees: \[150^{\circ} = \frac{5 \pi}{6}\]
So: \[2x = n \pi + x + \frac{5 \pi}{6}\]
wat the wrong in 1 st method!!
and subtract x from both sides: \[x = n \pi + \frac{5 \pi}{6}\]
My mistake in reading i read as : x - pie/3
thxxx
Welcome dear..

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