HELP with trigonometric identites (C3, chapter 7, edexcel)

Okay so I'm solving this equation: 2sin(x)-3cos(x)=1
I'm solving it by equating the left hand side to a single trigonometric equation.
What I did first: I equated the left hand side to a sine function. So
2sin(x)-3cos(x)=Rsin(x+a). Then I used the addition formula.
Rsin(x+a)= Rsin(x)*cos(a)+Rcos(x)*sin(a). Then I equated the coefficients.
So, Rcos(a)=2 and Rsin(a)=-3.
So I solved for R & a (not gonna go into the details) and found the R=sqrt(13) and a=-56.31. So 2sin(x)-3cos(x)=sqrt(13)sin(x-56.31)
Then I said sqrt(13)sin(x-56.31)=1 (as required in first equation), solved for x and got 72.4 and 220.2 (if 0<x<360). This, according to the text book solution, is the correct answer.
HERE COMES THE PROBLEM
As stated, I equated the left hand side to Rsin(x+a). But surely, it doesn't matter whether I equate the equation to Rsin(x+a), Rsin(x-a), Rcos(x+a) or Rcos(x-a). Surely, when I solve for x, it should be the same answer, regardless which formula I used. So I tested this out.
I equated 2sin(x)-3cos(x) to Rcos(x+a) instead
Expanded Rcos(x+a) to Rcos(x)cos(a)-Rsin(x)sin(a)
Equated coefficients: Rcos(a)=-3 & -Rsin(a)=2, so Rsin(a)=-2
Solved for R and a, and got R=sqrt(13) and a=33.69
So 2sin(x)-3cos(x)=sqrt(13)cos(x+33.69).
Set the equation: sqrt(13)cos(x+33.69)=1
Solved for x and got, *drum roll*, 40.21 and 252.41. Two completely different answers!!!!!!
Got so frustrated that I decided to generate the graphs on a graph sketcher to see what was going on. Turns out that sqrt(13)sin(x-56.31) generates the same graph as 2sin(x)-3cos(x), thus confirming that it is a correct equation.
However, the graph of sqrt(13)cos(x+36.69) generates the REFLECTION in the x axis, of 2sin(x)-3cos(x).
Why is this happening!!! What have I done wrong in calculating the cos equation to mean that it doesn't produce the same graph as the others!!!

i got the jist...

I suspect a common mistake occured

2sinx - 3cosx =Rsin(x - a) for the signs to agree

Yeah, it works with both the sine equations, Rsin(x-a) and Rsin(x+a), but it doesn't work with the cos ones. I mean, surely it shouldn't matter which one you use, as long as you equate the coefficients appropriately...
2sinx -3cosx can be a translation of EITHER cos(x) or sin(x). The problem is in calculating it

Yeah, it works with both the sine equations, Rsin(x-a) and Rsin(x+a), but it doesn't work with the cos ones. I mean, surely it shouldn't matter which one you use, as long as you equate the coefficients appropriately...
2sinx -3cosx can be a translation of EITHER cos(x) or sin(x). The problem is in calculating it

very difficult to explain on line but take a note of this

example 1

3cosx+4sinx

you can use Rsin(x+a) OR Rcos(x-a)

both expansions produce plus, so the signs match


example 2

3cosx-4sinx

you can use Rsin(x-a) OR Rcos(x+a)

both expansions produce minus, so the signs match again


I hope this helps

But do you know why it must, no question about it, be like this? Like why you can't equate a negative coefficient to a positive trig function, considering you expand appropriately and get the correct equation? Or is it just something I have to accept for now....