<s>[INST] Minimum value of trigonometric equation Find the minimum value of the expression $$y=\frac{16-8\sin^{2} 2x +8\cos^{4} x}{\sin^{2} 2x} .$$ When I convert the expression completely into $2x$, cross multiply and make the discriminant of the quadratic equation greater than $0$, I get the minimum value $-\infty$. I know it is wrong, but why?
 [/INST] $$8\cos^4x = 8\left(\dfrac{1+\cos (2x)}{2}\right)^2 = 8\left(\dfrac{1+2\cos (2x)+ \cos^2(2x)}{4}\right)=2+4\cos (2x)+2\cos^2(2x) = 2+4t+2t^2, t = \cos (2x) \Rightarrow y = \dfrac{16-8(1-t^2)+2+4t+2t^2}{1-t^2} = \dfrac{10+10t^2+4t}{1-t^2}=f(t), -1 \leq t \leq 1$$. Can you take it from here?
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