<s>[INST] Minima and maxima of a trig function let $f(x) = \sqrt {4\sin^4x - \sin^2x\cos^2x + 4\cos^4x}$ 
what is the product of the minimum and maximum value of the function?
I tried turning them all to $\sin x$
$-\sin^2x\cos^2x = \sin^4x-\sin^2x ; 4\cos^4x = (1-\sin^2x)^2 = 4sin^4x - 4sin^2x +4 $
which gives us :
$9\sin^4x - 2\sin^2x +4$
then i dont know how to proceed
EDIT: 
The equation above is wrong, i got if you try solving for the ones i have given : 
$9\sin^4x - 5\sin^2x +4$
then I tried making $sin^2x = a$ (Thanks @Greninja for the info)
where in the the min is when $a = 5/18$ and the value is 47/36, 
then the max is when $a=1$ and the value is 8 
Min x Max = $12\sqrt{94}$ which looks far fetched
I dont know where I went wrong with my solution Please help.
And yes, I have been told that the answer is $\sqrt{7}$
 [/INST] Use Indentity $$\sin^4 x+\cos^4 x = (\sin^2 x+\cos^2 x)^2-2\sin^2 x\cos^2 x = 1-2\sin^2 x\cos^2 x$$
So let $$y = \sqrt{4(\sin^4 x+\cos^4 x)-\sin^2 x\cos^2 x} = \sqrt{4-9\sin^2 x\cos^2 x} = \sqrt{4-\frac{9\sin^2(2x)}{4}}$$
Now use  $0 \leq \sin^2(2x)\leq 1$
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