1)
k = f ' (x0)
y ' = (2sinx - 3ctgx)' = 2cosx + 3/sin^2x
y ' (pi/3) = 2cospi/3 + 3/sin^2(pi/3) = 2*1/2 + 3/0.75 = 1 + 4 = 5
2)
k = f ' (x0)
y' = (cosx + 6tgx)' = - sinx + 6/cos^2x
y'(pi/6) = - sin pi/6 + 6/cos^2(pi/6) = - 1/2 + 6/0.75 = 7.5
3/4X5/7=15/28 3x5=15 4x7=28
12x+15y=-15 (-1)
12x+6y=12
-12x-15y=15
12x+6y=12
-15y=15
6y=12
9y=27
y=3
12x+15×3=-15
12x+45=-15
12x=60
x=5
Ответ: x=5; y=3