Given, tanθ=cot5θ
⇒tanθ=tan(π2−5θ)⇒π2−5θ=nπ+θ⇒6θ=π2−nπ⇒θ=π12−nπ6
Also cos4θ=sin2θ=cos(π2−2θ)
⇒4θ=2nπ±(π2−2θ)
Taking positive sign,
6θ=2nπ+π2⇒θ=nπ3+π12
Taking negative sign,
2θ=2nπ−π2⇒θ=nπ−π4
Also values of θ suggest that there are only 3 common solutions in the interval of (−π2,π2).