Question

1) 9^2+log₉2
2) 2cosx³+5cosx²-3cosx=0
3) log₃(7-x) ≥ 2log₃7

Answer 1

9^2+log₉2 =9²*9:^log(9)2=81*2=162

2(cosx)³+5(cosx)²-3cosx=0 
cosx=a
2a³+5a²-3a=0
a(2a²+5a-3)=0
a=0⇒cosx=0⇒x=π/2+πk,k∈z
2a²+5a-3=0
D=25+24=49
a1(-5-7)/4=-3⇒cosx=-3<-1 нет решения
а2=(-5+7)/4=1/2⇒cosx=1/2⇒x=+-π/6+2πk,k∈z

log₃(7-x) ≥ 2log₃7
log(3)(7-x)≥log(3)49
{7-x>0⇒x<7
{7-x≥49⇒x≤7-49⇒x≤-42
x∈(-∞;-42]