Sinx+sin2x=2cos^2x+cosx

Sinx+sin2x=2cos²x+cosx

sinx+2sinx*cosx=2cos²x+cosx

sinx+2sinx*cosx-2cos²x-cosx=0

(sinx+2sinxcosx) - (2cos²x+cosx) = 0

sinx (1+2cosx) - cosx (2cosx+1) = 0

sinx (2cosx+1) - cosx (2cosx+1) = 0

(sinx-cosx) (2cosx+1) = 0

sinx-cosx=0|:cosx≠0 или 2cosx+1=0

tgx-1=0 2cosx=-1|:2

tgx=1 cosx=-1/2

x=π/4+πn; n∈z x = + - 2π/3+2πn; n∈z

Ответ: π/4+πn; n∈z

+-2π/3+2πn; n∈z