$\begin{aligned} & 2 \sin ^{-1} x+3 \cos ^{-1} x=\frac{2 \pi}{5} \\ & \frac{\pi}{2}+\cos ^{-1} x=\frac{2 \pi}{5} \\ & \cos ^{-1} x=\frac{2 \pi}{5}-\frac{\pi}{2} \\ & \cos ^{-1} x=\frac{-\pi}{10} \end{aligned}$ Which is not possible as $\cos ^{-1} x \in[0, \pi]$ $\therefore \quad$ No solution