If ${\sin }^4\theta {\cos }^2\theta =\sum _{n=0}^{\infty }{a}_{2n}\cos 2n\theta ,$ then the least n for which $a_{2n}=0$ is

${\sin }^4\theta {\cos }^2\theta =\sum _{n=0}^{\infty }{a}_{2n}\cos 2n\theta$

$\Rightarrow {\left({\sin }^2\theta \right)}^2{\cos }^2\theta =\sum _{n=0}^{\infty }{a}_{2n}\cos 2n\theta$

$\Rightarrow {\left(1-{\cos }^2\theta \right)}^2{\cos }^2\theta =\sum a_{2n}\cos 2n\theta$

$\Rightarrow \left(1+{\cos }^4-2{\cos }^2\theta \right){\cos }^2\theta$

$=\sum a_{2n}\cos 2n\theta$

$\Rightarrow {\cos }^2\theta +{\cos }^4\theta .{\cos }^2\theta -2{\cos }^4\theta$

$=\sum a_{2n}\cos 2n\theta$

$\Rightarrow \left(\frac{1+\cos 2\theta }{2}\right)+{\left(\frac{1+\cos 2\theta }{2}\right)}^2\left(\frac{1+\cos 2\theta }{2}\right)$

$-2{\left(\frac{\cos 2\theta +1}{2}\right)}^2=\sum _{n=0}^{\infty }{a}_{2n}\cos 2n\theta$

On comparing both sides, we get minimum value of n is 1.