y' = -2cosx*sinx + cosx = 0
cosx(-2sinx + 1) = 0
cosx = 0 x = π/2 + πk, k ∈ Z
-2sinx + 1 = 0 sinx = 1/2 x = (-1)^k * π/6 + πk, k ∈ Z
Найдем значения x, принадлежащие промежутку [π/3;π]
x = π/2 + πk
при k = 0 x = π/2
x = (-1)^k * π/6 + πk
при k = 1; x = 5π/6
Проверим значния ф-ии в точках π/3; π/2; 5π/6 и π
y(π/3) = cos^2(π/3) + sin(π/3) = 1/4 + √3/2 = (2√3 + 1)/4 ≈ 1,11
y(π/2) = 0 + 1 = 1
y(5π/6) = 3/4 + 1/2 = 5/4 = 1,25
y(π) = 1 + 0 = 1
yнаиб = 1,25
yнаим = 1