Сколько корней имеет уравнение на отрезке [0; п/2]
2cos^2x - 1 = sin 3x

сos2x=sin3xsin3x-cos2x=0sin3x-sin(π/2-2x)=02sin(5x/2-π/4)cos(x/2+π/4)=0sin(5x/2-π/4)=0⇒5x/2-π/4=πn⇒5x/2=π/4+πn⇒x=π/10+2πn/5cos(x/2+π/4)=0⇒x/2+π/4=π/2+πn⇒x/2=π/4+πn⇒x=π/2+2πnx=π/10 U x=π/2