Question

Найдите корни уравнения 2sin3x=-1 на промежутке [-П;П] .

Answer 1

2sin3x=-1
sin3x=-1/2
3x=-π/6+2πk,k∈z⇒x=-π/18+2πk/3,k∈z
3x=-5π/6+2πk,k∈z⇒x=-5π/18+2πk/3,k∈z
-π≤-π/18+2πk/3≤π
-18≤-1+12k≤18
-17/12≤k≤19/12
k=-1⇒x=-π/18-2π/3=-13π/18
k=0⇒x=-π/18
k=1⇒x=-π/18+2π/3=11π/18

-π≤-5π/18+2πk/3≤π
-18≤-5+12k≤18
-13/12≤k≤23/12
k=-1⇒x=-5π/18-2π/3=-17π/18
k=0⇒x=-5π/18
k=1⇒x=-5π/18+2π/3=7π/18

Answer 2

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