Trigonometry question If $$\frac{3-\tan^2\frac{\pi}{7}}{1-\tan^2\frac{\pi}{7}}=\alpha \cos\frac{\pi}{7}.$$ If $\alpha$ is a natural number.Find $\alpha$.
My attempt is:
$$\frac{3-\tan^2\frac{\pi}{7}}{1-\tan^2\frac{\pi}{7}}=\alpha \cos\frac{\pi}{7}$$
convert it into sin,cos
$$\frac{3\cos^2\frac{\pi}{7}-\sin^2\frac{\pi}{7}}{\cos^2\frac{\pi}{7}-\sin^2\frac{\pi}{7}}=\alpha \cos\frac{\pi}{7}$$
$$\frac{3\cos^2\frac{\pi}{7}-\sin^2\frac{\pi}{7}}{\cos\frac{2\pi}{7}}=\alpha \cos\frac{\pi}{7}$$
$$3\cos^2\frac{\pi}{7}-\sin^2\frac{\pi}{7}=\alpha \cos\frac{\pi}{7}\cos\frac{2\pi}{7}$$
$$2\cos^2\frac{\pi}{7}+\cos^2\frac{\pi}{7}-\sin^2\frac{\pi}{7}=\alpha \cos\frac{\pi}{7}\cos\frac{2\pi}{7}$$
$$2\cos^2\frac{\pi}{7}+\cos\frac{2\pi}{7}=\alpha \cos\frac{\pi}{7}\cos\frac{2\pi}{7}$$
but i got stuck and could not further solve it....
I would appreciate the help,thanks in advance.
 A: $$\frac{3-\tan^2\frac{\pi}{7}}{1-\tan^2\frac{\pi}{7}}=1+\frac{2}{1-\tan^2\frac{\pi}7}=1+\frac{2\tan\frac{\pi}7}{1-\tan^2\frac{\pi}7}\frac1{\tan\frac{\pi}7}=1+\frac{\tan\frac{2\pi}{7}}{\tan\frac{\pi}7}\\=1+\frac{\sin\frac{2\pi}{7}\cos\frac{\pi}{7}}{\cos\frac{2\pi}{7}\sin\frac{\pi}{7}}=\frac{\sin\frac{3\pi}{7}}{\cos\frac{2\pi}{7}\sin\frac{\pi}{7}}=\frac{\sin\frac{4\pi}{7}}{\cos\frac{2\pi}{7}\sin\frac{\pi}{7}}=4\cos\frac{\pi}7.$$
A: Use $\tan^2A=\cdots=\dfrac{1-\cos^2A}{\cos^2A}$ and replace $\dfrac\pi7$ with $B$
to get $$2\alpha\cos^3B-4\cos^2B-\alpha\cos B+1=0\  \ \ \  (1)$$
Now if $7C=(2n+1)\pi$ where $n$ is any integer
$C=\dfrac{(2n+1)\pi}7$ where $n\equiv0,1,2,3\pmod7$
and $\cos4C=\cdots=-\cos3C\implies8\cos^4C-8\cos^2C+1=-(4\cos^3C-3\cos C)$
$\iff8\cos^4C+4\cos^3C-8\cos^2C-3\cos C+1=0\ \  \ \ (2)$ 
whose roots will be $\cos\dfrac{(2n+1)\pi}7$ where $n\equiv0,1,2,3\pmod7$
Now $n\equiv3\implies\cos\dfrac{(2n+1)\pi}7=\cdots=-1$
So, the roots of $0=\dfrac{8\cos^4C+4\cos^3C-8\cos^2C-3\cos C+1}{\cos C+1}=8\cos^3C-4\cos^2C-4\cos A+1\  \ \ \  (3)$ 
will be $\cos\dfrac{(2n+1)\pi}7$ where $n\equiv0,1,2\pmod7$
Now compare $(1),(3)$ to get $\dfrac{2\alpha}8=\dfrac44=\dfrac{\alpha}4=\dfrac11$
Observation: The given problem holds true for $\dfrac{(2n+1)\pi}7$ where $n\equiv0,1,2\pmod7$
$(3)$ can be derived like factor $z^7-1$ into linear and quadratic factors and prove that $ \cos(\pi/7) \cdot\cos(2\pi/7) \cdot\cos(3\pi/7)=1/8$
