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