$\tan 4x=\cot \left( \frac{\pi }{3}+4x \right)=\tan \left( \frac{\pi }{2}-\frac{\pi }{3}-4x \right)\Rightarrow \tan 4x=\tan \left( \frac{\pi }{6}-4x \right)\Rightarrow 4x=k\pi +\frac{\pi }{6}-4x\Rightarrow 8x=k\pi +\frac{\pi }{6}\Rightarrow x=\frac{k\pi }{8}+\frac{\pi }{48}$