A sad mistake students often do is take what their teacher says at face value, which is understandable, but it seems like you fell into the trap of not being taught this properly. It's always good to go over this topic from a different source to see if what you're doing is valid.
So you have that;
cosθ−sinθ=rsinθcosα−rcosθsinα.
Note that
θ is the main (and only) independent variable here. It varies.
r and
α do not. They are fixed in place, but it's your whole job in a question like this to determine what exactly they are fixed as.
To ensure both sides are equivalent, it is sufficient to ensure that the coefficients of
sinθ and
cosθ are the same.
For the coefficient of
cosθ, it is 1 on the LHS and
−rsinα on the RHS.
Therefore;
1=−rsinα.
For the coeff of
sinθ, it is -1 on the LHS and
rcosα on the RHS.
Therefore;
−1=rcosα.
It is now sufficient for you to determine what
r is (via Pythagorean theorem with the two results we just got) and also what
α is (there are infinitely many options for alpha, but this is why you are given the strict range of
0<α<2π in which there is only one value of alpha for you to have).
So have a go at finding these values.