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# ch- 3 trigonometry, ncert, ex-22

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solve tan 2x = - cot (x + pi/3)

naincy  Mar 22, 2017
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#1
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I have tried to post an answer but the site froze up and I lost it all.

Here is the short version

solve tan 2x = - cot (x + pi/3)

I solved by graphing.

Herer is the full graph

https://www.desmos.com/calculator/mblxlza8ms

From the graph it appears the answer is        $$x=n\pi+\frac{5\pi}{6}\qquad n \in Z$$

In my last (lost) answer I substituted this answer in to check it was correct and it was exactly so.

If you want to ask questions by all means do so :)

Here is the pic:

Melody  Mar 22, 2017
edited by Melody  Mar 22, 2017
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Don't know whether or not this is the long version, but Melody hasn't posted it so I will

$$\displaystyle \tan2x = -\cot(x+\pi/3),\\\frac{\sin2x}{\cos2x}=-\frac{\cos(x+\pi/3)}{\sin(x+\pi/3)},\\\sin2x\sin(x+\pi/3)=-\cos2x\cos(x+\pi/3),\\ \cos2x\cos(x+\pi/3)+\sin2x\sin(x+\pi/3)=0,\\\cos[2x-(x+\pi/3)]=0,\\\cos(x-\pi/3)=0,\\x-\pi/3=(2n+1)\pi/2=n\pi+\pi/2,\;\;n=0,\pm1,\pm2,\dots\\x=n\pi+5\pi/6,\;\;n=0,\pm1,\pm2,\dots$$

Guest Mar 23, 2017

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