решите уравнение sin2x+√2sinx=2cosx+√2

sin2x+√2sinx=2cosx+√22sinx*cosx+√2sinx=2cosx+√2sinx(2cosx+√2)-(2cosx+√2)=0(2cosx+√2)*(sin-1)=02cosx+√2=0  или sinx-1=01.2cosx=-√2cosx=-√2/2x=+-(π-arccos√2/2)+2πn, n∈Zx=+-3π/4+2πn, n∈Z2. sinx-1=0sinx=1x=π/2+2πn, n∈Z