# 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 $$\tan2A$$

We obtain: $$6\tan{A}(1-\tan^2A)=(1-\tan^2A)^2+4\tan^2A$$ or $$\frac{6\tan{A}}{1-\tan^2A}=1+\frac{4\tan^2A}{(1-\tan^2A)^2}$$ or $$\tan^22A-3\tan2A+1=0,$$ which gives $$\tan2A=\frac{3+\sqrt5}{2}$$ or $$\tan2A=\frac{3-\sqrt5}{3}.$$ Also, $$\tan^22A=\frac{1}{\cos^22A}-1<4-1=3,$$ which fives $$\tan2A=\frac{3-\sqrt5}{2}.$$
\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 $$t$$.