1
$\begingroup$

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 $\tan2A$

$\endgroup$
2
$\begingroup$

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 $$\tan2A=\frac{3-\sqrt5}{2}.$$

$\endgroup$
2
$\begingroup$

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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.