1
(2sinxcosx+4cosx)-(2√3sin²x+4√3sinx)=0
2cosx(sinx+2)-2√3sinx(sinx+2)=0
(sinx+2)(2cosx-2√3sinx)=0
sinx+2=0⇒sinx=-2<-1 нет решения<br>2cosx-2√3sinx=0/2cosx
1-√3tgx=0⇒tgx=1/√3⇒x=π/6+πk,k∈z
2
2sinxcosx+√3sinx=0
sinx(2cosx+√3)=0
sinx=0⇒x=πk,k∈z
cosx=-√3/2πx=+-π/6+2πk,k∈z
3
(tg³x+tg²x)-(3tgx+3)=0
tg²x(tgx+1)-3(tgx+1)=0
(tgx+1)(tg²x-3)=0
(tgx+1)(tgx-√3)(tgx+√3)=0
tgx+1=0tgx=-1x=-π/4+πk,k∈z
tgx-√3=0⇒tgx=√3⇒x=π/3+πk,k∈z
tgx+√3=0⇒tgx=-√3⇒x=-π/3+πk,k∈z