Question

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

Answer 1

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