F (x) = cos2x/cosx-sinx в точке x=П/4 помогите

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