Решите уравнение cos (pi/2+2x)=sqrt2*sinx

cos (pi/2+2x)=sqrt2*sinx-sin2x-sqrt2*sinx=02sinxcosx-sqrt2*sinx=0sqrt2*sinx*(sqrt2*cosx-1)=0sinx=0 cosx=sqrt2/2x=pi*n x=+- (pi/4)+2pi*n n-целое число [-5pi; -4pi]-5pi <= pi*n <= -4pi x=-5pi -5 <= n <= -4 n=-5,-4 x=-4pi -5pi <= pi/4+2pi*n <= -4pi-21pi/4 <= 2pi*n <= -17pi/4-21/8 <= n <= 17/8 -5pi <= -pi/4+2pi*n <= -4pi-19pi/4 <= 2pi*n <= -15pi/4-19/8 <= n <= -15/8