X²+2x+7>0
D1=k2-ac=4-7<0;корней нет
Решений нет
ОДЗ
{(x+3)/(x-3)>0⇒x<-3 U x>3
{x-3>0⇒x>3
{x+3>0⇒x>-3
x∈(3;∞)
перейдем к основанию 2
log(2)4/log(2)[(x+3)/(x-3)]=2(log(2)(x-3)/log(2)(1/2)-log(2)√(x+3)/log(2)(1/√2))
2/log(2)[(x+3)/(x-3)]=2(-log(2)(x-3)-1/2log(2)(x+3)/(-1/2))
2/log(2)[(x+3)/(x-3)]=2(log(2)[(x+3)/(x-3)]
log(2)[(x+3)/(x-3)]=t
2/t=2t
2t²=2
t²=1
t1=-1 U t2=1
log(2)[(x+3)/(x-3)]=-1
(x+3)/(x-3)=1/2⇒2x+6=x-3⇒x=-9∉ОДЗ
log(2)[(x+3)/(x-3)]=1
(x+3)/(x-3)=2⇒x+3=2x-6⇒x=9
Ответ х=9
5sin^2t - 9cost - 2 = 0
5*(1 - cos^2t) - 9cost - 3 = 0
-5cos^2t - 9cost + 2 = 0
5cos^2t + 9cost - 2 = 0
D = 81 + 4*5*2 = 121
1) cost = (-9 - 11)/10
cost = - 2 не удовлетворяет условию: I cost I ≤ 1
2) cost = (-9 + 11)/10
cost = 1/5
t = (+ -)arccos(1/5) + 2πn, n∈Z
Ответ: t = (+ -)arccos(1/5) + 2πn, n∈Z