In $\triangle PQR$, if $3\sin P+4\cos Q=6$ and $4\sin Q+3\cos P=1$, then the angle $R$ is equal to In $\triangle PQR$, if $3\sin P+4\cos Q=6$ and $4\sin Q+3\cos P=1$, then the angle $R$ is equal to 
My attempt is as follows:-
Squaring both equations and adding
$$9+16+24\sin(P+Q)=37$$
$$\sin(P+Q)=\dfrac{1}{2}$$
either $P+Q=\dfrac{\pi}{6}$ or $P+Q=\dfrac{5\pi}{6}$
If $P+Q=\dfrac{\pi}{6}$, then $R=\dfrac{5\pi}{6}$ otherwise $R=\dfrac{\pi}{6}$
Let's see case $1$: $P+Q=\dfrac{\pi}{6}$
$$3\sin P+4\cos\left(\dfrac{\pi}{6}-P\right)=6$$
$$3\sin P+4\left(\dfrac{\sqrt{3}}{2}\cos P+\dfrac{1}{2}\cdot\sin P\right)=6$$
$$3\sin P+2\sqrt{3}\cos P+2\sin P=6$$
$$5\sin P+2\sqrt{3}\cos P=6\tag{1}$$
$$4\left(\dfrac{1}{2}\cdot\cos P-\sin P\cdot\dfrac{\sqrt{3}}{2}\right)+3\cos P=1$$
$$-2\sqrt{3}\sin P+5\cos P=1\tag{2}$$
$$\cos P=\dfrac{12\sqrt{3}+5}{37}$$
$$\sin P=\dfrac{30-2\sqrt{3}}{37}$$
Using calculator I found $\cos P=0.69$, this means $P>\dfrac{\pi}{6}$ because $\cos \dfrac{\pi}{6}=0.866$, this mean $Q$ will be negative because $Q=\dfrac{\pi}{6}-P$. So this cannot be the case hence $P+Q$ would be $\dfrac{5\pi}{6}$ and $R$ will be $\dfrac{\pi}{6}$
This is the correct answer also , but I want to know does there exist any better way to decide on the value of $P+Q$. I am asking this because I had to use the calculator for finding the value of $\cos P$.

If $R=\frac{5\pi}6$, we have 
$$3\sin P + 4\cos(\frac\pi6-P)=6$$
Note that the RHS is an increasing function of $P$ for $P\in(0,\frac\pi6]$, whose maximal value is at $P=\frac\pi6$, i.e.
$$RHS_{max}=3\cdot \frac12+4\cdot 1 = 5.5 <6$$
Thus, $R\ne \frac{5\pi}6$.
