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