Найдите корни уравнения на заданном промежутке
Sin 3x= √2\2, [0;2Π]

3x=π/4+2πn U 3x=3π/4+2πkx=π/12+2πn/3 U x=π/4+2πk/30≤π/12+2πn/3≤2π0≤1+8n≤24-1≤8n≤23-1/8≤n≤23/8n=0⇒x=π/12n=1⇒x=π/12+2π/3=9π/12=3π/4n=2⇒x=π/12+4π/3=17π/120≤π/4+2πk/3≤2π0≤3+8k≤24-3≤8k≤21-3/8≤k≤21/8k=0⇒x=π/4k=1⇒x=π/4+2π/3=11π/12n=2⇒x=π/4+4π/3=19π/12