If $f(x) = \sin^4 x+\cos^2 x\;\forall x\; \in \mathbb{R}\;,$ Then $\bf{Max.}$ and $\bf{Min.}$ value of $f(x)$ 
If $f(x) = \sin^4 x+\cos^2 x\;\forall x\; \in \mathbb{R}\;,$ Then $\bf{Max.}$ and $\bf{Min.}$ value of $f(x).$

My Solution:: Let $$\displaystyle y = \sin^4 x+\cos^2 x \leq \sin^2 x+\cos^2 x=1$$
And for Minimum, We take $$\displaystyle y = \sin^4 x+\cos^2 x=(1-\cos^2 x)^2+\cos^2 x$$
So $$\displaystyle y=\cos^4 x-\cos^2 x+1 = \left(\cos^2 x-\frac{1}{2}\right)^2+\frac{3}{4}\geq \frac{3}{4}$$
So We get $\displaystyle y=\sin^4 x+\cos^2 x\in \left[\frac{3}{4}\;,1\right]$
My question is How can we find Min. of $f(x)$ other then that method, 
Something Like Using Inequality.,Thanks 

We have
\begin{align}
f(x) & = \sin^4(x) + \cos^2(x) = \sin^4(x) + 1 - \sin^2(x) = 1 - \sin^2(x)(1-\sin^2(x))\\
& = 1-\sin^2(x)\cos^2(x) = 1 - \dfrac{\sin^2(2x)}4
\end{align}
I trust you can finish off from here.
