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!!!