<span>(3x^2*y)^2*(4xy^3)^3=</span>3²x⁴y²4³x³y⁹=576x⁷y¹¹
<span>y = - 3tgx + 6x - 1,5П + 8 на отрезке [-П/3;П/3]
y`=-3/cos</span>²x+6=0
cos²x=1/2
(1+cos2x)/2=1/2
1+cos2x=1
cos2x=0
2x=π/2+πk
x=π/4+πk/2,k∈z
k=-1 x=-π/4∈[-π/3;π/3]
k=0 x=π/4∈[-π/3;π/3]
y(-π/3)=-3*(-√3)-2π-1,5π+8≈5-7,5+8=5,5 наибольшее
y(-π/4)=3-1,5π-1,5π+8≈11-9=2
y(π/4)=-3+1,5π-1,π+8=5
y(π/3)=-3√3+2π-1,5π+8≈4,4
Решение задания приложено
2cos(x-п/2)-1=2cos(-(п/2-x))-1=2cos(п/2-x)-1=2cos(x)-1