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lululilovo
@lululilovo
August 2022
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Срочноооо, срочнооооо, Срооооочнооо. С 22 по 24. Буду благодарна
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oganesbagoyan
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Решение: см. фото
22. Найдите область определения функции : y =√(1 - 2cos2x)
---
1 - 2cos2x ≥ 0 ⇔cos2x ≤ 1/2 ⇒2πn + π/3 ≤ 2x ≤ 2π - π/3 + 2πn ,n ∈ Z. ⇔
π/3+2πn ≤ 2x ≤ 5π/3+2πn,n ∈ Z. ⇔ π/6+ πn ≤ x ≤ 5π/6+ πn , n ∈ Z.
или иначе x ∈ [ π/6 + πn ; 5π/6+ πn ] , n ∈ Z.
ответ : C)
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Answers & Comments
Verified answer
Решение: см. фото22. Найдите область определения функции : y =√(1 - 2cos2x)
---
1 - 2cos2x ≥ 0 ⇔cos2x ≤ 1/2 ⇒2πn + π/3 ≤ 2x ≤ 2π - π/3 + 2πn ,n ∈ Z. ⇔
π/3+2πn ≤ 2x ≤ 5π/3+2πn,n ∈ Z. ⇔ π/6+ πn ≤ x ≤ 5π/6+ πn , n ∈ Z.
или иначе x ∈ [ π/6 + πn ; 5π/6+ πn ] , n ∈ Z.
ответ : C)