4. Solve the equation $\sqrt[4]{1-\cos ^{7} 15 x \cos ^{2} 9 x}=\sin 9 x . \quad$ (8 points)
Solution: Given the condition $\sin 9 x \geq 0$, both sides of this equation can be squared. Under the found constraints, the equation is equivalent to the following: $\sin ^{4} 9 x+\cos ^{7} 15 x \cos ^{2} 9 x-1=0,\left(1-\cos ^{2} 9 x\right)^{2}+\cos ^{7} 15 x \cos ^{2} 9 x-1=0$, $-2 \cos ^{2} 9 x+\cos ^{4} 9 x+\cos ^{7} 15 x \cos ^{2} 9 x=0, \cos ^{2} 9 x\left(\cos ^{2} 9 x+\cos ^{7} 15 x-2\right)=0$. Thus, we arrive at the system of equations:

1) $\cos 9 x=0, x=\frac{\pi}{18}+\frac{\pi n}{9}, n \in Z$, taking into account the condition $\sin 9 x \geq 0$ we have $x=\frac{\pi}{18}+\frac{2 \pi n}{9}, n \in Z$.
2) $\cos ^{2} 9 x+\cos ^{7} 15 x-2=0$, which is equivalent to the system of equations $\left\{\begin{array}{l}\cos ^{2} 9 x=1, \\ \cos ^{7} 15 x=1,\end{array} \Leftrightarrow\right.$

$\left\{\begin{array}{c}9 x=\pi k, \\ 15 x=2 \pi m,\end{array} k, m \in Z\right.$. Therefore, $5 k=6 m, k, m \in Z \Rightarrow m=5 s, k=6 s, s \in Z$
$\Rightarrow x=\frac{2 \pi s}{3}, s \in Z$, which satisfies the condition $\sin 9 x \geq 0$.

Thus, the solutions to the original equation are: $x=\frac{\pi}{18}+\frac{2 \pi n}{9}, n \in Z, x=\frac{2 \pi s}{3}, s \in Z$.

Answer: $x=\frac{\pi}{18}+\frac{2 \pi n}{9}, n \in Z, x=\frac{2 \pi s}{3}, s \in Z$.