Найдите наибольшее значение функции y=cos2x+√2x-√2/2 на отрезке [/4:/2]

Y`=-2sin2x+√2=0
2sin2x=√2
sin2x=√2/2
1)2x=π/4+2πn,n∈Z
x=π/8∉[π/4;π/2]
2)2x=3π/4+2πk,k∈Z
x=3π/8∈[π/4;π/2]
y(π/4)=2cosπ/2+π√2/4-√2/2=√2/4*(π-2)≈0,4 наибольшее
y(3π/8)=2cos3π/4+3π√2/4-√2/2=-√2/2+3π√2/4-√2/2=√2/4(3π-4)≈-0,3
y(π/2)=2cosπ+π√2/2-√2/2=-2+√2/2(π-1)≈-2+1,5≈-0,5