4. Given $\sin ^{10} x+\cos ^{10} x=\frac{11}{36}$. Then $\sin ^{14} x+\cos ^{14} x=$ $\qquad$
4. $\frac{41}{216}$.

Let $\sin ^{n} x+\cos ^{n} x=a_{n}, \sin ^{2} x \cdot \cos ^{2} x=y \leqslant \frac{1}{4}$.
Then $a_{n+2}=a_{n}-y a_{n-2}(n \geqslant 2)$, where,
$$
\begin{array}{l}
a_{0}=2, a_{2}=1, a_{4}=1-2 y, a_{6}=1-3 y, \\
a_{10}=5 y^{2}-5 y+1=\frac{11}{36} .
\end{array}
$$

Solving for $y$ gives $y=\frac{1}{6}$.
Thus, $\sin ^{14} x+\cos ^{14} x=a_{14}=\frac{41}{216}$.