the expression can be rewritten as $1/2(a(1+\cos2x)+c(1-\cos2x)+b\sin2x) =(1/2)((a+c)+(a-c)\cos2x+b\sin2x)$. Rest is simple

Submathematics wrote: the expression can be rewritten as $1/2(a(1+\cos2x)+c(1-\cos2x)+b\sin2x) $ $=(1/2)((a+c)+(a-c)\cos2x+b\sin2x)$. Rest is simple $P=acos{^2}x+bsinxcosx+csin{^2}x$ $=\frac{1}{2}\left((a+c)+(a-c)\cos2x+b\sin2x\right)$, $P_{max}=\frac{1}{2}\left((a+c)+\sqrt{(a-c)^2+b^2}\right)$, $P_{min}=\frac{1}{2}\left((a+c)-\sqrt{(a-c)^2+b^2}\right)$.

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