2. Solve the inequality $2 \cdot 5^{2 x} \cdot \sin 2 x-3^{x} \geq 5^{2 x}-2 \cdot 3^{x} \cdot \sin 2 x$.
Solution. Transform the inequality:

$$
5^{2 x}(2 \sin 2 x-1) \geq-3^{2 x}(2 \sin 2 x-1) \Rightarrow\left(5^{2 x}+3^{2 x}\right)(2 \sin 2 x-1) \geq 0
$$

Note that $5^{2 x}+3^{2 x}>0$ for any $x \in \mathbf{R}$, so the given inequality is equivalent to the inequality $(2 \sin 2 x-1) \geq 0$. Solving this, we find $\frac{\pi}{12}+2 \pi k \leq x \leq \frac{5 \pi}{12}+2 \pi k, k \in \mathbf{Z}$.

Answer: $\frac{\pi}{12}+2 \pi k \leq x \leq \frac{5 \pi}{12}+2 \pi k, k \in \mathbf{Z}$.