(tg²x-3)√(11cosx)=0 x∈[5π/2;-π]
либо tg²x-3=0, либо cosx=0
1. tg²x-3=0
tg²x=3
tgx=+-√3
x=+-π/3+πn
+-π/3+πn=-5π/2
+-1/3+n=-5/2
n=5/2+-1/3=(-15+-2)/6=-17/6 -13/6 → n=-2
+-π/3+πn=-π
+-1/3+n=-1
n=-1+-1/3 =-4/3 -2/3 →n=-1
x=+-π/3-2π=-(6+-1)π/3; x=-7π/3 x=-5π/3
x=+-π/3-π=-(3+-1)π/3; x=-4π/3 (x=-2π/3 не попадает в интервал)
2. cosx=0
x=π/2+πn
x=-3π/2
x=-5π/2
Ответ: -5π/2; -7π/3; -5π/3; -3π/2; -4π/3;