3
2√3sinxcosx+cos²x-sin²x-2sin²x-2cos²x=0
3sin²x-2√3sinxcosx+cos²x=0/cos²x
3tg²x-2√3tgx+1=0
(√3tgx-1)²=0
√3tgx=1
tgx=1/√3
x=π/6+πk,k∈z
4
[(1-cos2x)²/4-(1-cosx)/2]:(1+co6x)=
=[(1-2cos2x+cos²2x)/4-(1-cos2x)/2]:(1+cos6x)=
=(1-2cos2x+cos²2x-2+2cos2x)/4:(1+cos6x)=
=(cos²2x-1)/4:(1+cos6x)=-sin²2x/(1+cos6x)
sin²2x/(1+cos6x)=0
sin²2x=0,cos6x≠-1
2x=πk⇒x=πk/2,k∈z
5
(tg5x+tgx)/(1-tg5xtgx)*(√3sin²x+sinx)=0
tg(5x+x)*sinx*(√3sinx+1)=0
tg6x=0⇒6x=πk⇒x=πk/6.k∈z
sinx=0⇒x=πk,k∈z
sinx=-1/√3⇒x=(-1)^(k+1)*arcsin1/√3+πk,k∈z
ответ x=πk/6.k∈z,x=(-1)^(k+1)*arcsin1/√3+πk,k∈z