<s>[INST] Let $\frac{1}{2}<\cos2A<1$ and $6\tan A-6\tan^3A=\tan^4A+2\tan^2A+1$, find $\tan 2A$ Let $\dfrac{1}{2}<\cos2A<1$ and $6\tan A-6\tan^3A=\tan^4A+2\tan^2A+1$, find $\tan 2A$
My attempt:
\begin{align*}
6\tan A(1-\tan^2A)&=\tan^4A+2\tan^2A+1\\
12\tan^2A&=\tan2A\tan^4A+2\tan2A\tan^2A+\tan2A\\
0&=\tan2A(\tan^4A)+(2\tan2A-12)\tan^2A+\tan2A\\
\because\tan2A&\in\mathbb{R}\\
\therefore \tan2A&\leqslant3
\end{align*}
From $\dfrac{1}{2}<\cos2A<1$ gives $0\leqslant\tan2A<\sqrt{3}$
Alfter using 2 inequality, I still can't find the exact value of $\tan2A$
 [/INST] Note that the given eqn can be reduced to : 
$\begin{align} & 6\tan A (1- \tan ^2 A) =(1+\tan^2A)^2 \\ \implies & 6\sin A \cos 2A \sec^3A=\sec^4A \\ \implies & 6\sin A\cos 2A\cos A=1\\ & \implies 3\sin 2A \cos 2A=1\\ & \implies \sin4A=2/3\\ & \implies \frac{2t}{1+t^2}=2/3 \\\end {align} $, where $t=\tan2A$. Solve for $t$.  Can you take it from here?
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