Given, $$\tan \theta = \cot 5\theta $$

$$ \Rightarrow \tan \theta = \tan \left( {{\pi \over 2} - 5\theta } \right)$$

$$ \Rightarrow {\pi \over 2} - 5\theta = n\pi + \theta $$

$$ \Rightarrow 6\theta = {\pi \over 2} - {{n\pi } \over 6}$$

$$ \Rightarrow \theta = {\pi \over {12}} - {{n\pi } \over 6}$$

Also $$\cos 4\theta = \sin 2\theta = \cos \left( {{\pi \over 2} - 2\theta } \right)$$

$$ \Rightarrow 4\theta = 2n\pi \, \pm \,\left( {{\pi \over 2} - 2\theta } \right)$$

Taking positive

$$6\theta = 2n\pi + {\pi \over 2} \Rightarrow \theta = {{n\pi } \over 3} + {\pi \over {12}}$$

Taking negative

$$2\theta = 2n\pi - {\pi \over 2} \Rightarrow \theta = n\pi - {\pi \over 4}$$

Above values of $$\theta$$ suggests that there are only 3 common solutions.