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 

$$\cos^4x-\cos^2x=\cos^2x(\cos^2x-1)=-\sin^2x\cos^2x=-\dfrac{(2\sin x\cos x)^2}4=-\dfrac{\sin^22x}4$$
For real $x,0\le\sin^22x\le1\iff-1\ge-\sin^22x\ge0$
In case you don't know double angle formula, $(\sin x\pm\cos x)^2\ge0\iff-1\le2\sin x\cos x\le1$
