4. Solve the equation $\sin 7 x+\sqrt[4]{1-\cos ^{11} 3 x \cos ^{2} 7 x}=0 . \quad(8$ points)
Solution: Given the condition $\sin 7 x \leq 0$, both sides of this equation can be squared. Under the found constraints, the equation is equivalent to the following: $\sin ^{4} 7 x+\cos ^{11} 3 x \cos ^{2} 7 x-1=0$, $\left(1-\cos ^{2} 7 x\right)^{2}+\cos ^{11} 3 x \cos ^{2} 7 x-1=0,-2 \cos ^{2} 7 x+\cos ^{4} 7 x+\cos ^{11} 3 x \cos ^{2} 7 x=0$, $\cos ^{2} 7 x\left(\cos ^{2} 7 x+\cos ^{11} 3 x-2\right)=0$. Thus, we arrive at the system of equations:

1) $\cos 7 x=0, x=\frac{\pi}{14}+\frac{\pi n}{7}, n \in Z$, taking into account the condition $\sin 7 x \leq 0$ we have $x=-\frac{\pi}{14}+\frac{2 \pi n}{7}, n \in Z$.
2) $\cos ^{2} 7 x+\cos ^{11} 3 x-2=0$, which is equivalent to the system of equations $\left\{\begin{array}{l}\cos ^{2} 7 x=1, \\ \cos ^{11} 3 x=1,\end{array} \Leftrightarrow\left\{\begin{array}{c}7 x=\pi k, \\ 3 x=2 \pi m\end{array}\right.\right.$ $k, m \in Z$. Therefore, $3 k=14 m, k, m \in Z \Rightarrow m=3 s, k=14 s, s \in Z \Rightarrow x=2 \pi s, s \in Z$, which satisfies the condition $\sin 7 x \leq 0$.

Thus, the solutions to the original equation are: $x=-\frac{\pi}{14}+\frac{2 \pi n}{7}, n \in Z \quad x=2 \pi s, s \in Z$.

Answer: $x=-\frac{\pi}{14}+\frac{2 \pi n}{7}, n \in Z \quad x=2 \pi s, s \in Z$.