2. It is known that the equation $3 \sin x+4 \cos x=a$ has exactly two equal real roots $\alpha, \beta$ in the interval $(0,2 \pi)$, $\alpha+\beta=$ $\qquad$ .
2. $\pi-2 \arcsin \frac{4}{5}$ or $3 \pi-2 \arcsin \frac{4}{5}$.

The equation can be transformed into $\sin (x+\varphi)=\frac{a}{5}, \varphi=\arcsin \frac{4}{5}$. From the given, we have $\left\{\begin{array}{l}\sin (\alpha+\varphi)=\frac{a}{5} \\ \sin (\beta+\varphi)=\frac{a}{5}\end{array} \Rightarrow \sin (\alpha+\varphi)-\right.$ $\sin (\beta+\varphi)=0 \Rightarrow 2 \cos \frac{\alpha+\beta+2 \varphi}{2} \sin \frac{\alpha-\beta}{2}=0$. Since $\alpha \neq \beta, \alpha, \beta \in(0,2 \pi)$, we know $\sin \frac{\alpha-\beta}{2} \neq 0$, thus $\cos \left(\frac{\alpha+\beta+2 \varphi}{2}\right)=0$. And $\varphi<\frac{\alpha+\beta+2 \varphi}{2}<2 \pi+\varphi<\frac{5 \pi}{2}$, so $\frac{\alpha+\beta+2 \varphi}{2}=\frac{\pi}{2}$ or $\frac{3 \pi}{2}$. Therefore, $\alpha+\beta=\pi-2 \arcsin \frac{4}{5}$ or $\alpha+\beta=3 \pi-2 \arcsin \frac{4}{5}$.