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НоВыЙ13
@НоВыЙ13
November 2021
1
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Решите неравенство:
[tex] cos^{2} 2x - sin^{2} 2x \geq \frac{1}{2} [/tex]
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Ranbu
Cos^2 2x-sin^2 2x≥1/2
cos4x≥1/2
-pi/3+2pik≤4x≤pi/6+2pik
-pi/12+pik/2≤x≤pi/12+pik/2
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Thanks 1
НоВыЙ13
а откуда взялось pi/6+2pik?
Ranbu
ой,прошу прощения,там pi/3
НоВыЙ13
это значит ответ будет ----- Х принадлежит [-pi/12+pik/2; pi/12+pik/2] ???
Ranbu
да
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Answers & Comments
cos4x≥1/2
-pi/3+2pik≤4x≤pi/6+2pik
-pi/12+pik/2≤x≤pi/12+pik/2