Question

Find the value of $x$ for the given equation

$\tan 2x=-\cot (x+\frac{\pi }{3})$

Answer 1

$\because \tan ({90}^0+\theta )=-\cot \theta$

$\Rightarrow \cot \theta =-\tan (90+\theta )$

$\cot \theta =-\tan (\frac{\pi }{2}+\theta )$

Replacing $\theta$ by $x+\pi /3$

$\Rightarrow -\cot (x+\frac{\pi }{3})=\tan (\frac{\pi }{2}+x+\frac{\pi }{3})$

Now given $\tan \left(2x\right)=-\cot (x+\pi /3)$

$\Rightarrow \tan 2x=\tan (\frac{\pi }{2}+\frac{\pi }{3}+x)$

$\Rightarrow \tan 2x=\tan (\frac{3\pi +2\pi }{2\times 3}+x)$

$\Rightarrow \tan 2x=\tan (\frac{5\pi }{6}+x)\to \left(1\right)$

General solution

$tanx=tany$

$\Rightarrow \tan 2x=\tan 2y\to \left(2\right)$

From $\left(1\right)$ & $\left(2\right)$

$\tan 2y=\tan (\frac{5\pi }{6}+x)$

$\Rightarrow 2y=\frac{5\pi }{6}+x$

General solution is

$2x=n\pi +2y$ where $n\in z$

put $2y=(x+\frac{5\pi }{6})$

$\Rightarrow 2x=n\pi +(x+\frac{5\pi }{6})$

$\Rightarrow 2x-x=n\pi +\frac{5\pi }{6}$

$\Rightarrow x=n\pi +\frac{5\pi }{6},n\in z$