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ТКУиК9ks11
@ТКУиК9ks11
August 2022
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Решите неравенство пожалуйста))) 1/2sin(x/3) > 1/4
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армения20171
1/2sinx/3>1/4
sinx/3>1/4:1/2
sinx/3>1/2
π/6+2πk<x/3<(π-π/6)+2πk
π/6+2πk<x/3<5π/6+2πk;k€Z
π/2+6πk<x<5π/2+6πk;k€Z
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Answers & Comments
sinx/3>1/4:1/2
sinx/3>1/2
π/6+2πk<x/3<(π-π/6)+2πk
π/6+2πk<x/3<5π/6+2πk;k€Z
π/2+6πk<x<5π/2+6πk;k€Z