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ellabcgbvc
@ellabcgbvc
July 2022
2
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решите неравенства |cosx|>или =√3/2
|tgx|>или =√3/3
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Бахит
Раскрываем скобки модуля с + и с -
-cosx≥√3/2 cosx≥√3/2
cosx≤-√3/2 x≥π/6+2πR
x≤5π/6+2πR
-tgx≥√3/3 tgx≥√3/3
tgx≤√3/3 x≥π/6+πR
x≤π/6+πR
2 votes
Thanks 1
Uturna
Verified answer
1) cosx≥√3/2,
cosx≤-√3/2
x∈[-π/6+πk; π/6+πk], k∈Z
2) tgx≥√3/3
tgx≤-√3/3
x∈(-π/2+πк;-π/6+πк]U[π/6+πk; π/2+πk), k∈Z
1 votes
Thanks 1
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Answers & Comments
-cosx≥√3/2 cosx≥√3/2
cosx≤-√3/2 x≥π/6+2πR
x≤5π/6+2πR
-tgx≥√3/3 tgx≥√3/3
tgx≤√3/3 x≥π/6+πR
x≤π/6+πR
Verified answer
1) cosx≥√3/2,cosx≤-√3/2
x∈[-π/6+πk; π/6+πk], k∈Z
2) tgx≥√3/3
tgx≤-√3/3
x∈(-π/2+πк;-π/6+πк]U[π/6+πk; π/2+πk), k∈Z