Question

cos2x=3(cos³x-sin³x)
Найдите корень уравнения.

Answer 1

cos2x=3(cos³x-sin³x)
cos²x-sin²x=3(cosx-sinx)(cos²x+sinxcosx+sin²x)

(cosx-sinx)(cosx+sinx-3-3sinxcosx)=0

1)cosx-sinx=0/:cosx≠0
1-tgx=0
tgx=1;x=π/4+πk;k€Z

2)cosx+sinx-3sinxcosx-3=0
√2*sin(π/4+x)-3/2sin2x-3=0

x€∅
x=π/4;sin(π/4+π/4)=1
sin2x=1

[sin2x=-1 ;x=-π/4+πk
[sin(x+π/4)=-1;x+π/4=-π/2+2πk;x=-π/4-π/2+2πk
x=-3π/4+2πk