Question

найдите максимум функции f(x)=cos(2018x)+sin(2018x)

Answer 1

$f(x)=cos(2018x)+sin(2018x)=cos(y)+sin(y)=\\ =[cos(y)*frac{sqrt{2}}{2}+sin(y)*frac{sqrt{2}}{2}]*frac{2}{sqrt{2}}=\\ =[cos(y)*cos(frac{pi}{4})+sin(y)*sin(frac{pi}{4})]*sqrt{2}=\\ =cos(y-frac{pi}{4})*sqrt{2}=sqrt(2)*cos(2018x-frac{pi}{4})\\ -1 leq cos(2018x-frac{pi}{4}) leq 1\\ -sqrt{2} leq sqrt{2}*cos(2018x-frac{pi}{4}) leq sqrt{2}\\ -sqrt{2} leq f(x) leq sqrt{2}$

Максимального значения $sqrt{2}$ функция $f(x)$ достигает при условии $cos(2018x-frac{pi}{4})=1$

$2018x-frac{pi}{4}=2pi n, nin Z\\ 2018x=frac{pi}{4}+2pi n, nin Z\\ x=frac{pi}{4*2018}+frac{2pi n}{2018}, nin Z\\ x=frac{pi}{8072}+frac{pi n}{1009}, nin Z$
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множество точек  $(frac{pi}{8072}+frac{pi n}{1009}; sqrt{2}), nin Z$ есть максимумами функции $f(x)$