Question

Решить уравнения 1. cosx+cos2x+cos3x=0 2. cos2x=cosx-sinx

Answer 1

Решение смотри на фотографии

Answer 2

1)2cos2xcosx+cos2x=0
cos2x(2cosx+1)=0
cos2x=0⇒2x=π/2+πn⇒x=π/4+πn/2
cosx=-1/2⇒x=+-2π/3+2πn
2)cos²x-sin²x-(cosx-sinx)=0
(cosx-sinx)(cosx+sinx)-(cosx-sinx)=0
(cosx-sinx)(cosx+sinx-1)=0
1)cosx-sinx=0
sin(π/2-x)-sinx=0
2sin(π/4-x)cosπ/4=0
sin(π/4-x)=0
π/4-x=πn
x=π/4+πn
2)cosx+sinx-1=0
cosx+sinx=1
sin(π/2-x)+sinx=1
2sinπ/4cos(π/4-x)=1
√2cos(π/4-x)=1
cos(π/4-x)=√2/2
π/4-x=π/4+2πn  U  π/4-x=-π/4+2πn
x=2πn  U  x=π/2+2πn