2 cos^2 x - 3sgrt3*cos x + 3 < 0; Сначала найдем корни этого неравенства, потом решим само неравенство методом интервалов. 2 cos^2 x - 3sgrt3*cos x + 3 = 0; D = (3sgrt3)^2 - 4*2*3 = 9*3 - 24 = 3= (sgrt3)^2; cos x = (3sgrt3 + sgrt3)/4 = 4sgrt3/4= sgrt3 или cos x = (3sgrt3 - sgrt3)/4 = 2sgrt3/4= sgrt3/2;
2(cos x - sgrt3)(cos x - sgrt3/2) <0
+ - + sgrt3/2sgrt3/2 cos x
sgrt3/2 < cos x < 1, так как -1<=cos x <=1; - pi/6 + 2pi*k < x < pi/6 + 2pi*k; k-Z
2(3х+2у)+9=4х+21
2х+10=3-(6х+5у)
2(3х+2у)+9=4х+21
2х+10=3-6х-5у
8х=-5у-7
2(3х+2у)+9=4х+21
х=-0,625у-0,875
6х+4у+9=4х+21
х=-0,625у-0,875
2х + 4у=12
х=-0,625у-0,875
2(-0,625у-0,875) + 4у=12
х=-0,625у-0,875
-1,25у +4у =12+ 1,75
х=-0,625у-0,875
2,75у= 13,75
у= 5
х= -0,625*5-0,875
у=5
х=-4
ответ: (-4;5)