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

2sin3x=-1 sin3x=-1/2 3x=-п/6+2пn,n€z x=-п/18+2пn/3,n€z 3x=7п/6+2пn,n€z x=7п/18+2пn/3,n€z -п≤-п/18+2пn/3≤п -п+п/18≤2пn/3≤п+п/18 |:п -1+1/18≤2n/3≤1+1/18 -17/18≤2n/3≤19/18 |*3 -17/6≤2n≤19/6 |:2 -17/12≤n≤19/12 при n=0 x=-п/18 при n=-1 x=-п/18-2п/3=-п/18-12п/18=-13п/18 при n=1 x= -п/18+2п/3=-п/18+12п/18=11п/18 -п≤7п/18+2пn/3≤п |:п -1≤7/18+2n/3≤1 -1-7/18≤2n/3≤1-7/18 -25/18≤2n/3≤11/18 |*3 -25/6≤2n≤11/6 |:2 -25/12≤n≤11/12 при n=-2 x=7п/18-4п/3=7п/18-24п/18=-17п/18 при n=-1 x=7п/18-2п/3=7п/18-12п/18=-5п/18 при n=0 x=7п/18