how to find values of x for which a triangle exists Given a $\triangle PQR , QR=3 , PR= x , PQ=2x$  and $\angle PQR= \theta$  calculate the values of $x$ for which the triangle exists.  I don't even know where to start.

The answer given by @Deepak is undoubtly the shortest possible and the one which is certainly awaited at your level.
Nevertheless, here is an answer that gives another proof for the result: 
$$1<x<3$$
in the framework of a "global understanding" (see figure below). 
It involves a classical result : the locus of points $P$ such that the ratio of their distances to two fixed points is constant (here $\dfrac{PQ}{PR}=2=\dfrac{2x}{x}$) is a circle (https://www.cut-the-knot.org/Curriculum/Geometry/LocusCircle.shtml)
(see commentary below) 
(more generally, constant 2 can be any positive constant).
On the figure, you see a triangle $PQR$ where $RQ$ is fixed and $P$ can be any point on the circle.
Now take the extreme points :


*

*$P$ is in $P_1$ iff $x=1$,

*$P$ is in $P_2$ iff $x=3$,
all values of $x$ in the range $[1,3]$ being allowed, with the restriction that these extreme points give flat triangles...
Commentary: this circle has its center on line $RQ$, with diameter $P_1P_2$ where $P_1$ and $P_2$ are the two extreme points on line $RQ$ mentionned above.

