7. Given the equation $8 t^{3}-4 t^{2}-4 t+1=0$ has a root $x$ in $\left(0, \frac{\pi}{13}\right)$, then $x=$
7. $\sin \frac{\pi}{14}$

Analysis: Note that: $8 x^{3}-4 x^{2}-4 x+1=0 \Rightarrow (x+1)\left(8 x^{3}-4 x^{2}-4 x+1\right)=0$, i.e., $3 x-4 x^{3}=8 x^{4}-8 x+1$
Also, $\cos 4 x=8 \sin ^{4} x-8 \sin x+1, \sin 3 x=3 \sin x-4 \sin ^{3} x, \frac{\pi}{13}\frac{\pi}{13}$, so $a \in\left(0, \frac{\pi}{6}\right)$, thus $a=\frac{\pi}{14}$; i.e., $x=\sin \frac{\pi}{14}$;