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dashadomrachev
@dashadomrachev
July 2022
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Решить неравенство корень из 3/2cosx+1/2sinx<=корень из 2/2
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shevtsovavalia
√3/2cosx+1/2sinx≤√2/2
cosπ/6cosx+sinπ/6sinx≤√2/2
cos(π/6-x)≤√2/2
π/4+2πn≤π/6-x≤7π/4+2πn
π/12+2πn≤-x≤19π/12+2πn
-19π/12-2πn≤x≤-π/12-2πn
Ответ: [-19π/12-2πn;-π/12-2πn]
3 votes
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Answers & Comments
cosπ/6cosx+sinπ/6sinx≤√2/2
cos(π/6-x)≤√2/2
π/4+2πn≤π/6-x≤7π/4+2πn
π/12+2πn≤-x≤19π/12+2πn
-19π/12-2πn≤x≤-π/12-2πn
Ответ: [-19π/12-2πn;-π/12-2πn]