Trigonometry with multiple angle and exact value of $\tan\pi/5$ By considering the equation $\tan5\theta=0$, show that the exact value of $\tan\pi/5$ is $\sqrt{5-2\sqrt{5}}$. 
Do I need to evaluate the multiple angle for $\tan5\theta=0$?

$$\tan(5x) = \dfrac{\tan(3x) + \tan(2x)}{1-\tan(3x) \tan(2x)} = \dfrac{\dfrac{3 \tan(x) - \tan^3(x)}{1-3 \tan^2(x)} + \dfrac{2 \tan(x)}{1-\tan^2(x)}}{1- \dfrac{3 \tan(x) - \tan^3(x)}{1-3 \tan^2(x)} \cdot \dfrac{2 \tan(x)}{1-\tan^2(x)}}$$
Hence,
$$\tan(5x) = 0 \implies (3t-t^3)(1-t^2) + 2t(1-3t^2) = t \left(t^4-4t^2+3 - 6t^2+2 \right) = 0$$
where $t= \tan(x)$. Since we are interested in $\tan(\pi/5)$, we can rule out $t=0$. Hence, we need to solve a bi-quadratic $t^4-10t^2+5 = 0$. This gives us
$$t^2 = \dfrac{10 \pm \sqrt{100-4 \times 5}}{2} = 5 \pm 2 \sqrt5 \implies t = \pm \sqrt{5 \pm 2\sqrt5}$$ We also know that
$\tan(\pi/5) \in (\tan(0),\tan(\pi/4)) = \left(0,1 \right)$. Hence, $$\tan(\pi/5) = \sqrt{5 - 2\sqrt5}$$
