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f(x) = Asin^{2}(x) + Bcos^{2}(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?

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# How to find minimum turning points

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