Write $\alpha$ for the two equal angles in the isosceles triangle, so that the angle at the apex is $\pi - 2\alpha$. By the sine rule, $$\frac{AB}{AC} = \frac{\sin\alpha}{\sin(\pi - 2\alpha)} = \frac{\sin\alpha}{\sin( 2\alpha)} = \frac{\sin\alpha}{2\sin\alpha\cos\alpha} = \frac1{2\cos\alpha}.$$ Now for a bit of trigonometry: $$\begin{aligned} \sin(3\theta) = \sin(2\theta+\theta) &= \sin(2\theta)\cos\theta + \cos(2\theta)\sin\theta \\ &= 2\sin\theta\cos^2\theta + \cos(2\theta)\sin\theta = \sin\theta(2\cos^2\theta + \cos(2\theta)) = \sin\theta(1 + 2\cos(2\theta)) \end{aligned}$$ (because $2\cos^2\theta = \cos(2\theta) + 1$). Therefore $1+ 2\cos(2\theta) = \dfrac{\sin(3\theta)}{\sin\theta}.$ In particular, with $\theta = \frac\pi7$, $$1 + 2\cos\tfrac{2\pi}7 = \frac{\sin\frac{3\pi}7}{\sin\frac{\pi}7} = \frac{\sin\frac{3\pi}7}{\sin\frac{6\pi}7} = \frac{\sin\frac{3\pi}7}{2\sin\frac{3\pi}7\cos\frac{3\pi}7} = \frac1{2\cos\frac{3\pi}7}.$$ It follows that if $$\frac{AB}{AC} = 1 + 2\cos\tfrac{2\pi}7$$ then $$ \frac1{2\cos\alpha} = \frac1{2\cos\frac{3\pi}7}$$, so that $\alpha = \frac{3\pi}7$. Thus the angles of the triangle are $\frac{3\pi}7$, $\frac{3\pi}7$ and $\frac\pi7$.