Выражение: 1)3*cos2x-3/tg2x+1-2 2)5tg2x*cosx+5sin2x*ctg2x

1)3·cos2x-3/tg2x+1-2=

    ((12·cos(x)^3-8·cos(x))·sin(x)-6·cos(x)^2+3)/(2·cos(x)·sin(x))

2)5tg2x·cosx+5sin2x·ctg2x=

(10·cos(x)^2·sin(x)+20·cos(x)^4-20·cos(x)^2+5)/(2·cos(x)^2-1)

1)3*cos2x-3/tg2x+1-2=((12*cos(x)^3-8*cos(x))*sin(x)-6*cos(x)^2+3)/(2*cos(x)*sin(x))

2)5tg2x*cosx+5sin2x*ctg2x=(10*cos(x)^2*sin(x)+20*cos(x)^4-20*cos(x)^2+5)/(2*cos(x)^2-1)

1)   (sin(x))^2-0,5*sin(2x)=0

(sin(x)^2)-0,5*2*sin(x)*cos(x)=0

sin(x)*(sin(x)-cos(x))=0

a)   sin(x)=0

x=pi*n

б) sin(x)-cos(x)=0

sin(x)/cos(x)=1

tg(x)=1

x=pi/4+pi*n

ответ:   x=pi*n

=pi/4+pi*n

2)   sqrt(2)*(cos(x)^2)+cos(x)-sqrt(2)=0

cos(x)=t

sqrt(2)*t^2+t-sqrt(2)=0

t1,2=(-1+-sqrt(1+8)/(2*sqrt(2)

a)   t1=-2/sqrt(2)

cos(x)=-2/sqrt(2)> 0 - не удовлетворяет одз

б)   t2=1/sqrt(2)

cos(x)=1/sqrt(2)

x=+-pi/4+2*pi*n

ответ:   x=+-pi/4+2*pi*n