Given $A={\sin }^2\theta +{\cos }^4\theta$, then for all real $\theta$

We have,

$\begin{array}{l}A={\sin }^2\theta +{\cos }^4\theta \\ \Rightarrow A=\frac{1-\cos 2\theta }{2}+{\left(\frac{1+\cos 2\theta }{2}\right)}^2\\ \Rightarrow A=\frac{1}{2}-\frac{1}{2}\cos 2\theta +\frac{1}{4}+\frac{1}{2}\cos 2\theta +\frac{1}{4}{\cos }^{2}2\theta \end{array}$

$\begin{array}{l}\Rightarrow A=\frac{3}{4}+\frac{1}{4}\left\{\frac{\cos 4\theta +1}{2}\right\}=\frac{3}{4}+\frac{1}{8}+\frac{1}{8}\cos 4\theta \\ \text{ Now, }-1\le \cos 4\theta \le 1\\ \Rightarrow \frac{-1}{8}\le \frac{\cos 4\theta }{8}\le \frac{1}{8}\\ \Rightarrow \frac{3}{4}+\frac{1}{8}-\frac{1}{8}\le \frac{3}{4}+\frac{1}{4}\left\{\frac{1+\cos 4\theta }{2}\right\}\le \frac{3}{4}+\frac{1}{8}+\frac{1}{8}\\ \Rightarrow \frac{3}{4}\le A\le 1\end{array}$