2. Solve the equation $\sin ^{4}(2025 x)+\cos ^{2019}(2016 x) \cdot \cos ^{2018}(2025 x)=1$.

(12 points)
# Solution:

$\sin ^{4} 2025 x+\cos ^{2019} 2016 x \cdot \cos ^{2018} 2025 x=1 \Leftrightarrow\left(1-\cos ^{2} 2025 x\right)^{2}+\cos ^{2019} 2016 x \cdot \cos ^{2018} 2025 x=1$

$\Leftrightarrow \cos ^{4} 2025 x-2 \cos ^{2} 2025 x+\cos ^{2019} 2016 x \cdot \cos ^{2018} 2025 x=0 \Leftrightarrow$

$\cos ^{2} 2025 x\left(\cos ^{2} 2025 x+\cos ^{2019} 2016 x \cdot \cos ^{2016} 2025 x-2\right)=0 \Leftrightarrow$

1) $\cos 2025 x=0, x=\frac{\pi}{4050}+\frac{\pi n}{2025}, n \in Z$ or 2) $\cos ^{2} 2025 x+\cos ^{2019} 2016 x \cdot \cos ^{2016} 2025 x=2 \Leftrightarrow$ $\left\{\begin{array}{c}\cos ^{2} 2025 x=1, \\ \cos ^{2019} 2016 x \cdot \cos ^{2016} 2025 x=1,\end{array} \Leftrightarrow\left\{\begin{array}{l}\sin 2025 x=0, \\ \cos 2016 x=1,\end{array} \Leftrightarrow\left\{\begin{array}{l}x=\frac{\pi n}{2025}, n \in Z, \\ x=\frac{2 \pi m}{2016}, m \in Z,\end{array}\right.\right.\right.$ $\frac{n}{2025}=\frac{m}{1008}, n=\frac{2025 m}{1008}=\frac{225 m}{112}, n \in Z, m \in Z \Rightarrow m=112 k, \quad x=\frac{\pi k}{9}, k \in Z$.

Answer: $x=\frac{\pi}{4050}+\frac{\pi n}{2025}, n \in Z, x=\frac{\pi k}{9}, k \in Z$.