Ответ:(π/6+πk;π/3+πk),k∈z
Объяснение: sinx·cosx>√3/4;
2sinx·cosx>√3/2;
sin2x>√3/2;
π/3+2πk<2x<2π/3+2πk,k∈z;
π/6+πk<x<π/3+πk,k∈z.
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Ответ:(π/6+πk;π/3+πk),k∈z
Объяснение: sinx·cosx>√3/4;
2sinx·cosx>√3/2;
sin2x>√3/2;
π/3+2πk<2x<2π/3+2πk,k∈z;
π/6+πk<x<π/3+πk,k∈z.