12sin²x(3ππ/2+x)=cosx2cos²x-cosx=0cosx(2cosx-1)=0cosx=0⇒x=π/2+πn,n∈z-3π/2≤π/2+πn≤0-3≤1+2n≤0-4≤2n≤-1-2≤n≤-1/2n=-2⇒x=π/2-2π=-3π/2n=-1⇒x=π/2-π=-π/22cosx-1=0cosx=1/2⇒x=+-π/3+2πn,n∈z-3π/2≤-π/3+2πn≤0-9≤-2+12n≤0-7≤12n≤2-7/12≤n≤2/12n=0⇒x=-π/3-3π/2≤π/3+2πn≤0-9≤2+12n≤0-11≤12n≤-2-11/12≤n≤-2/12 нет решения2tg³x+tg²x-3tgx-3=0tg²(tgx+1)-3(tgx+1)=0(tgx+1)(tgx-√3)(tgx+√3)=0tgx+1=0⇒tgx=-1⇒x=-π/4+πn,n∈z2π≤-π/4+πn≤7π/28≤-1+4n≤149≤4n≤159/4≤n≤15/4n=3⇒x=-π/4+3π=11π/4tgx-√3=0⇒tgx=√3⇒x=π/3+πn,n∈z2π≤π/3+πn≤7π/212≤2+6n≤2110≤6n≤1910/6≤n≤19/6n=2⇒x=π/3+2π=7π/3n=3⇒x=π/3+3π=10π/3tgx+√3=0⇒tgx=-√3⇒x=-π/3+πn,n∈z2π≤-π/3+πn≤7π/212≤-2+6n≤2114≤6n≤2314/6≤n≤23/6n=3⇒x=-π/3+3π=8π/3