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

sin2x+√2sinx=2cosx+√2
2sinx*cosx+√2sinx=2cosx+√2
sinx(2cosx+√2)-(2cosx+√2)=0
(2cosx+√2)*(sin-1)=0
2cosx+√2=0  или sinx-1=0

1.2cosx=-√2
cosx=-√2/2
x=+-(π-arccos√2/2)+2πn, n∈Z
x=+-3π/4+2πn, n∈Z

2. sinx-1=0
sinx=1
x=π/2+2πn, n∈Z