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

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2sin3x=-1sin3x=-1/23x=-π/6+2πk,k∈z⇒x=-π/18+2πk/3,k∈z3x=-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≤19/12k=-1⇒x=-π/18-2π/3=-13π/18k=0⇒x=-π/18k=1⇒x=-π/18+2π/3=11π/18-π≤-5π/18+2πk/3≤π-18≤-5+12k≤18-13/12≤k≤23/12k=-1⇒x=-5π/18-2π/3=-17π/18k=0⇒x=-5π/18k=1⇒x=-5π/18+2π/3=7π/18

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