F'(x)=(9sin^2x)' - (3cos4x)'= 18cosx*sinx + 12sin4x
(x - 9)^2 - (x - 3)^2 = 0
( x - 9 - (x - 3)) * ( x - 9 + x - 3) = 0
- 6 * ( 2x - 12) = 0
2x - 12 = 0
x - 6 = 0
x = 6
<span>COS(A+П/4)*COS(A-П/4) = (1/2)* [cos(</span>α + π/4 - α + π/4) + cos(α + π/4 + α - π/4)] =
= (1/2)* [ cos(π/2) + cos2α] = (1/2)*cos(2α)
Эта система уравнений из задания 3 не имеет ни одного решения