I have a function: f(x) = Asin2(x) + Bcos2(x) + Csin(2x) and I want to find the minimum turning point(s). To start with, I calculated: f'(x) = (A - B)sin(2x) + 2Ccos(2x) Therefore, turning points occur when f'(x)=0, or: tan(2x) = 2C / (B - A) To find the minima, I then want to look at the 2nd derivative: f''(x) = 2(A - B)cos(2x) - 4Csin(2x) So, as far as I understand, minima will occur when both f'(x)=0 and f''(x)>0. So the second inequality can be expressed as: 2(A - B)cos(2x) > 4Csin(2x) Now I'm having trouble knowing how to proceed. Since sin and cos can be positive or negative depending on x, I can't see a neat way of really understanding where the minima will be. Could anyone help?