Решите и распишите как решать​

1)3-log(3) x=log(3) (x-6)
ОДЗ х > 0 и х - 6>0
x > 6
log(3) 27 - log(3) x = log(3) (x - 6)
log(3) (27/x) = log(3) (x - 6)
27/x = x - 6
27 = x^2 - 6x
x^2 - 6x - 27 = 0
D = 36 + 108 = 144
x1 = (6+12)/2 = 18/2 = 9
x2 = (6 - 12)/2 = -6/2 = -3 - лишний корень
ответ 9
2)log(7) (5x+4)≤2
ОДЗ 5х + 4 > 0
x > -0.8
log(7) (5x + 4) ≤ log(7) 49
5x + 4≤ 49
5x ≤ 45
x ≤ 9
-0.89
ответ x ∈(-0.8; 9]
3){log4x-log4 y=1/2
{3x-y=10
ОДЗ x>0 и y>0
y = 10 - 3x
log(4) x - log(4) (3x - 10) = 1/2
log(4) (x/(3x - 10)) = log(4) √4
x/(3x - 10) = 2
x = 2(3x- 10)
x = 6x - 20
-5x = -20
x = 4
y = 3*4 - 10 = 12 - 10 = 2
ответ: (4; 2)

1)tg²x+ctg²x+3tgx+3cygx=-4
(tgx+ctgx)²-2tgxctgx+3(tgx+ctgx)+4=0      tgx*ctgx=1
(tgx+ctgx)²+3(tgx+ctgx)+2=0
tgx+ctgx=a
a²+3a+2=0
a1+a2=-3 U a1*a2=2
a1=-2⇒tgx+ctgx=-2
tgx+1/tgx+2=0
tg²x+2tgx+1=0  tgx≠0
(tgx+1)²=0⇒tgx=-1⇒x=-π/4+πn
a2=-1⇒tgx+ctgx=-1
tgx+1/tgx+1=0
tg²x+tgx+1=0  tgx≠0
tgx=t⇒t²+t+1=0
D=1-4=-3-решения нет
2)(1-сos2x)/2+(1-cos4x)/2=(1+cos6x)/2+(1+cos8x)/2
1-cos2x+1-cos4x=1+cos6x+1+cos8x
cos8x+cos6x+cos4x+cos2x=0
2cos5xcos3x+2cos5xcosx=0
2cos5x(cos3x+cosx)=0
2cos5x*2cos2xcosx=0
4cos5xcos2xcosx=0
cos5x=0⇒5x=π/2+πn⇒x=π/10+πn/5
cos2x=0⇒2x=π/2+πn⇒x=π/4+πn/2
cosx=0⇒x=π/2+πn