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 $\cos P$.
 A: A solution using complex numbers geometry :
The 2 relationships can be grouped into a single one by adding the second one to $i$ times the first one, giving : 
$$3e^{iP}+4ie^{-iQ}=1+6i \ \ \iff \ \ \underbrace{3e^{iP}}_A+\underbrace{4e^{i(\pi/2-Q)}}_B=\underbrace{1+6i}_C \tag{1}$$
This defining relationship between points ("affixes") of these complex numbers can be written under a vectorial form : 
$$\vec{OA}+\vec{OB}=\vec{OC}$$
meaning that $OBCA$ is a parallelogram with prescribed lengths $OA, OB, OC$ which will not leave much degrees of freedom as we are going to see it. 
Remark: the polar angles of $\vec{OA}$ and $\vec{OB}$ are $P$ and $\pi/2-Q$ resp. (the latter being in $(-\pi/2,\pi/2)$). 
Therefore
$$\alpha := angle(OB,OA)=P-(\pi/2-Q)\tag{2}$$

In parallelogram $OBCA$, we have the following classical relationship between the sides and the diagonals (see here).
$$p^2+q^2=2(a^2+b^2)\tag{3}$$
With $a=OA=3, b=OB=4, p=OC=\sqrt{1^2+6^2}=\sqrt{37}$, we deduce from (3) that the second diagonal has its length $q$ given by :
$$37+q^2=2(3^2+4^2) \ \ \implies \ \ AB=q=\sqrt{13}$$
Let us apply the cosine formula to triangle $OAB$ :
$$AB^2=OA^2+OB^2-2OA.OB \cos \alpha \ \ \iff \ \ 13=3^2+4^2-2.3.4 \cos \alpha$$
giving 
$$\cos(OB,OA)=\cos \alpha = \dfrac12 \ \ \implies \ \ \alpha := angle(OB,OA)=\dfrac{\pi}{3}\tag{4}$$
Identifying (4) and (2), we get : 
$$P+Q=\dfrac{\pi}{2}+\dfrac{\pi}{3}=\dfrac{5\pi}{6} \ \ \implies  \ \ R=\pi-(P+Q)=\dfrac{\pi}{6}$$
as awaited. 
But there is a case we haven't yet considered :
It has been assumed implicitly that the polar angle of $\vec{OB}$ is less than the polar angle of $\vec{OA}$. We could have had the inverse situation, which geometrically corresponds to a symmetry of parallelogram $OBCA$ with respect to its diagonal $OC$. Fortunately, this cannot arise, because the polar angle for $\vec{OB}$ would have been outside $(-\pi/2,\pi/2)$, contradicting the remark done upwards.  
A: 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, $R\ne \frac{5\pi}6$.
