Since this must be true for all values of θ, and cos and sin are distinct functions, no non-zero multiple of cosθ could ever be equal to 4 sinθ for all values of θ. Therefore, the overall multiple of cosθ on the left-hand-side must be 0.
Therefore, Acosθ + Bcosθ ≡ 0and (A+B) cosθ ≡ 0so A = - B
We can then plug this back into the equation to solve for A:A(sinθ + cosθ) - A(cosθ - sinθ) ≡ 4sinθAsinθ - (-Asinθ) = 4sinθ [the cosθ terms cancel one another out]Asinθ + Asinθ = 4sinθ2A = 4A = 2B = - A = - 2