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