1
tgx=2tgx/(1-tg²x)
{tgx-tg³x-2tgx=0⇒tg³x+tgx=0⇒tgx(tg²x+1)=0⇒tgx=0⇒x=πn,n∈z (tg²x+1>0)
{1-tg²x≠0
Ответ x=πn,n∈z
2
2sinxcosx-2√3cos²x=0
2cosx(sinx-√3cosx)=0
cosx=0⇒x=π/2+πn,n∈z
sinx-√3cosx=0/cosx⇒tgx-√3=0⇒tgx=√3⇒x=π/3+πn,n∈z
3
sin²x-2cosx=0
1-cos²x-2cosx=0
cosx=a
a²+2a-1=0
D=4+4=8
a1=(-2-2√2)/2=-1-√2⇒cosx=-1-⇒2<-1 нет решения<br>a2=-1+√2⇒cosx=√2-1⇒x=+-arccos(√2-1)+2πn,n∈z
4
2cos(5x/2)*cos(3x/2)=0
cos(5x/2)=0⇒5x/2=π/2+πn⇒x=π/5+2πn/5,n∈z
cos(3x/2)=0⇒3x/2=π/2+πn⇒x=π/3+2πn/3,n∈z
5
2*1-3cosx-2=0
-3cosx=0
cosx=0
x=π/2+πn,n∈z
6
2sin²x-sinx=0
sinx(2sinx-1)=0
sinx=0⇒x=πn,n∈z
sinx=1/2⇒x=(-1)^n*π/6+πn,n∈z