11. (15 points) Given the equation $8 t^{3}-4 t^{2}-4 t+1=$ 0 has a root $x$ in $\left(0, \frac{\pi}{13}\right)$. Find $x$.


11. Notice
$$
\begin{array}{l}
8 x^{3}-4 x^{2}-4 x+1=0\left(x \in\left(0, \frac{\pi}{13}\right)\right) \\
\Rightarrow(x+1)\left(8 x^{3}-4 x^{2}-4 x+1\right)=0 \\
\Rightarrow 8 x^{4}+4 x^{3}-8 x^{2}-3 x+1=0 \\
\Rightarrow 3 x-4 x^{3}=8 x^{4}-8 x+1 .
\end{array}
$$

For any $\alpha \in \mathbf{R}$, we have
$$
\begin{array}{l}
\sin 3 \alpha=3 \sin \alpha-4 \sin ^{3} \alpha, \\
\cos 4 \alpha=8 \sin ^{4} \alpha-8 \sin ^{2} \alpha+1 .
\end{array}
$$

Given $x \in\left(0, \frac{\pi}{13}\right)$, let $x=\sin \alpha$. Then
$$
\sin 3 \alpha=\cos 4 \alpha \text {. }
$$

Since $x \in\left(0, \frac{\pi}{13}\right)$ and $\sin \frac{\pi}{6}=\frac{1}{2}>\frac{\pi}{13}$, by the monotonicity of $\sin x$, we must have
$$
\alpha \in\left(0, \frac{\pi}{6}\right), 3 \alpha, 4 \alpha \in\left(0, \frac{\pi}{2}\right) .
$$

Therefore, $3 \alpha+4 \alpha=7 \alpha=\frac{\pi}{2}$, which means $\alpha=\frac{\pi}{14}$. Hence, $x=\sin \frac{\pi}{14}$.