Question

cosx*log(12)(9-x^2)=0

Answer 1

ОДз
(3-x)(3+x>0
x=3  x=-3
x∈(-3;3)
cosx=0⇒x=π/2+πn,n∈z
x1=-π/2
x2=π/2
log(12)(9-x²)=0
9-x²=1
8-x²=0
x²=8
x=-2√2
х=2√2
x={-π/2;-2√2;2√2;π/2}