$$max(\sin 3A)=1\\\max(\cos 4B)=1 \\\\\sin 3A+\cos 4B=1+1 \to \\\simultaniusly \\\ \begin{cases}\sin 3A=1 & A\leq\pi \to 3A\leq 3\pi \\\cos 4B=1 & B\leq \pi \to 4B\leq 4\pi\end{cases} $$ $$\begin{cases}\sin 3A=1 & 3A= \dfrac{\pi}{2},\dfrac{5\pi}{2} \\\cos 4B=1 & 4B=0,2\pi,4\pi\end{cases}$$ in second one $4B=0,4\pi$ are not acceptable because $A+B+C=180 \to B\
eq0 ,180^0$ so $4B=2\pi \to B=\dfrac{\pi}{2}$ $$\hat B=90 \to \begin{cases} 3A= \dfrac{\pi}{2}=90 \to \hat A=30 \to \hat C=60 \\\3A= \dfrac{5\pi}{2}\to A=150 \to A+B+C=90+150+C =180 (no-solution)\end{cases}$$