1
2sin²2x+sin²4x=5/4
2(1-cos4x)/2+(1-cos8x)/2=5/4
4-4cos4x+2-2cos8x-5=0
1-4cos4x-4cos²4x+2=0
cos4x=a
4a²+4a-3=0
D=16+48=64
a1=(-4-8)/8=-1,5⇒cos4x=-1,5<-1 нет решения<br>a2=(-4+8)/8=1/2⇒cos4x=1/2⇒4x=+-π/3+2πn⇒x=+-π/12+πn/2,n∈z
2
1/(cos2xcos3x)+1/(cos3xcos4x)+1/(cos4xcos5x)=0
1/(cos2xcos3x)+1/(cos3xcos4x)=(cos4x+cos2x)/(cos2xcos3xcos4x)=
=2cos3xcosx)/(cos2xcos3xcos4x)=2cosx/(cos2xcos4x),cos3x≠0
2cosx/(cos2xcos4x)+1/(cos4xcos5x)=0
2cosxcos5x+cos2x=0,cos2x≠0 u cos4x≠0 U cos5x≠0
2*1/2*(cos4x+cos6x)+cos2x=0
cos4x+cos6x+cos2x=0
cos4x+2cos4xcos2x=0
cos4x*(1+2cos2x)=0cos4x=0 не удов усл
cos2x=-1/2
2x=+-2π/3+2πn
x=+-π/3+πn,n∈z
3
(1-2tgx)/(1-tgx)=(1-2sinx/cosx):(1-sinx/cosx)=(cosx-2sinx)/(cosx-sinx)
cosx-sinx=(cosx-2sinx)/(cosx-sinx)
(cos²x-2sinxcosx+cos²x)=(cosx-2sinx),sinx≠cosx
1-2sinxcosx-cosx+2sinx=0
(1-cosx)+2sinx(1-cosx)=0
(1-cosx)(1+2sinx)=0
cosx=1⇒x=2πn,n∈z
sinx=-1/2⇒x=(-1)^(n+1)*π/6+πn,n∈z