Cos³x=cos²x/2sinx
2sinxcos³x-cos²x=0,sinx≠0
cos²x(2sinxcosx-1)=0
cosx=0⇒x=π/2+πn,n∈z
-π≤π/2+πn≤-π/2
-2≤1+2n≤-1
-3≤2n≤-2
-3/2≤n≤-1
n=-1⇒x=π/2-π=-π/2
2sinxcosx=1
sin2x=1⇒2x=π/2+2πk,k∈z⇒x=π/4+πk,k∈z
-π≤π/4+πk≤-π/2
-4≤1+4k≤-2
-5≤4k≤-3
-5/4≤k≤-3/4
k=-1⇒x=π/4-π=-3π/4
π/2ж-3π/4Ъ
Ответ:
f(1)<g(7)
Объяснение:
f(x)=x²+1 f(1)=1²+1=2
g(x)=x²-1 g(7)=7²-1=49-1=48
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