2sin^2 x-sin2x+sinx=cosx на отрезке [0;3п/2]

2sin^2x-sin2x+sinx-cosx=02sin^2x-2sinx*cosx+sinx-cosx=0-2sinx*cosx-cosx+2sin^2x+sinx=0-cosx(2sinx+1)+sinx(2sinx+1)=0(2sinx+1)(-cosx+sinx)=01) 2sinx+1=02sinx=-1sinx=-1/2x=-π/6+2πn, n∈Zx=-5π/6+2πn, n∈Z2) sinx=cosx/:cosx≠0sinx/cosx=1tgx=1x=π/4+πk, k∈Z