Question

(√3+sin(2x)+3cos(x)) / (1+2sin(x))= -√3+cos(x)

Answer 1

ОДЗ
1+2sinx≠0⇒2sinx≠-1⇒sinx≠-1/2⇒x≠-π/6+2πk U x≠-5π/6+πk,k∈z
√3+sin2x+3cosx=(1+2sinx)(cosx-√3)
√3+sin2x+3cosx=cosx-√3+2sinxcosx-2√3sinx
√3+sin2x+3cosx-cosx+√3-sin2x+2√3sinx=0
2√3+2cosx+2√3sinx=0 разделим на 2
√3+cosx+√3sinx=0
√3sin²(x/2)+√3*cos²(x/2)+cos²(x/2)-sin²(x/2)+2√3sin(x/2)cos(x/2)=0
(√3-1)*sin²(x/2)+2√3sin(x/2)cos(x/2)+(√3+1)*cos²(x/2)=0
разделим на cos²(x/2)
(√3-1)*tg²(x/2)+2√3*tg(x/2)+(√3+1)=0
введем новую переменную
a=tg(x/2)
(√3-1)*a²+2√3*a+(√3+1)
D=(2√3)²-4*(√3-1)(√3+1)=12-4*(3-1)=12-4*2=12-8=4,√D=2
a1=(-2√3-2)/(2√3-2)=-(√3+1)/(√3-1)=-(√3+1)(√3+1)/[(√3-1)(√3+1)=
=-(3+2√3+1)/(3-1)=-2(2+√3)/2=-(2+√3)
tg(x/2)=-(2+√3)⇒x/2=-arctg(2+√3)+πk⇒x=-2arctg(2+√3)+2πk,k∈z
a2=(-2√3+2)/(2√3-2)=-1
tg(x/2)=-1⇒x/2=-π/4+πk⇒x=-π/2+2πk,k∈z