The function $f\left(x\right)={\sin }^{2}x{\cos }^{3}x$  attains a maximum when $x=$ 

$\text{ Let, }\begin{array}{l}y={\sin }^{2}x{\cos }^{3}x\\ \therefore y^{\mathrm{\prime }}=2\sin x\cos x{\cos }^{3}x+{\sin }^{2}x3{\cos }^{2}x(-\sin x)\\ or,y^{\mathrm{\prime }}=2\sin x{\cos }^{4}x-3{\sin }^{3}x{\cos }^{2}x\end{array}$

$$\begin{gathered} \text{ for maximum or minimum value }{y}^{\mathrm{\prime }}=0 \\ \therefore \sin x{\cos }^{2}x\left(2{\cos }^{2}x-3{\sin }^{2}x\right)=0 \\ \Rightarrow \sin x{\cos }^{2}x=0 or 2{\cos }^{2}x-3{\sin }^{2}x=0 \\ \begin{array}{r}\Rightarrow \frac{{\cos }^{2}x}{{\sin }^{2}x}=\frac{3}{2}\\ \Rightarrow \tan x=\sqrt{\frac{2}{3}}\end{array} \end{gathered}$$

Maximum at $x=Ta{n}^{-1}\sqrt{\frac{m}{n}}=Ta{n}^{-1}\sqrt{\frac{2}{3}}$