Cos(6x) = cos^2(3x) - sin^2(3x)
sin(6x) = 2sin(3x)*cos(3x)
cos^2(3x) - sin^2(3x) - 2√3*sin(3x)*cos(3x) = -cos^2(3x) - sin^2(3x)
2cos^2(3x) - 2√3*sin(3x)*cos(3x) = 0
2cos(3x)*(cos(3x) - √3*sin(3x)) = 0
1) cos(3x) = 0, 3x = π/2 + πk, x = π/6 + πk/3
2) √3*sin(3x) = cos(3x), tg(3x) = 1/√3, 3x = π/6 + πk, x = π/18 + πk/3
О,Ваше задание решено!Ответ во вложении!!!
1)f(x)=cos(x^3)
d/du [cos (u)]d/dx [x^3]
-sin(u)d/dx [x^3]
-sin(x^3)d/dx [x^3]
-sin(x^3)(3x^2)
-3sin(x^3)x^2
-3x^2 sin(x^3)