HELP with trigonometric identites

[Deep_Thinker97]

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 going to 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 textbook 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!

 

[Mark44]

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 going to 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 textbook 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!

cos(x) = -sin(x - 90°), which you can verify by expanding the right side.

So cos(x + 33.69°) = -sin(x - 90° + 33.69°) = -sin(x + 56.31°)
So the left and right sides are equal in magnitude, but opposite in sign. Adding a sign to sin(x + 56.31°) causes a reflection across the x-axis.