Question

Solve $tan2x=-cot(x+\frac{\pi }{3})$

Answer 1

$(\sqrt{3}-1)\cos \theta +(\sqrt{3}+1)\sin \theta =2$ $\to \left(1\right)$

Put $\sqrt{3}=1=r\sin \alpha$ and $\sqrt{3}+1=r\cos \alpha$

$r^2=(\sqrt{3}+1{)}^2+(\sqrt{3}+1{)}^2\Rightarrow r^2=8\Rightarrow r=2\sqrt{2}$

Also $\tan \alpha =\frac{\sqrt{3}-1}{\sqrt{3}+1}=2-\sqrt{3}\Rightarrow \alpha =\frac{\pi }{12}$

From $\left(1\right)$ we have

$r\sin \alpha \cos \theta +r\cos \alpha \sin \theta =2$

$r\sin (\theta +\alpha )=2$

$\sin (\theta +\alpha )=\frac{1}{\sqrt{2}}\cdot \sin (\theta +\alpha )=\sin \frac{\pi }{4}$

$\theta +\alpha =n\pi +(-1{)}^n\frac{\pi }{4}$

$\theta =n\pi +(-1{)}^n\frac{\pi }{4}-\frac{\pi }{12}$ $nRz$.