А)Int(3/[sin(3x)]^2-cos(2x)) dx = Int 3/[sin(3x)]^2 dx - Int cos(2x) dx =
Int d(3x)/[sin(3x)]^2 - 0.5*Int cos(2x) d(2x) = -ctg(3x) -0.5*sin(2x) + C.
-ctg(-3*pi/2)-0.5*sin(-2*pi/2)+C=3
0 0 C=3
Answer: -ctg(3x) -0.5*sin(2x)+3
Б)2Pin, n прин. Z x = +-(Pi - arccos(1/3)) + 2Pin, n прин. Z 3) sqrt(3) * cosx = sinx |:cosx tgx = sqrt(3) x = arctg(sqrt(3)) + Pik, k прин. Z x = Pi/3 + Pik, k прин
Точно не уверенa!!!