Question

Даю 50 баллов
Помогите пожалуйста с уравнением

Answer 1

$frac{2-cos4x+(4-sqrt3), sin2x+2, cos2x-2sqrt3}{1+2sin(frac{9pi }{2}+2x)}=1\\star ; sin(frac{9pi }{2}+2x)=sin(4pi +frac{pi }{2}+2x)=sin(frac{pi }{2}+2x)=cos2x; ; star \\\ODZ:; 1+2cos2xne 0; ,; cos2xne -frac{1}{2}; ,; 2xne pm arccos(-frac{1}{2})+2pi n\\2xne pm (pi -frac{pi }{3})+2pi n; ,; 2xne pm frac{2pi }{3}+2pi n; ,; underline {xne pm frac{pi }{3}+pi n; ,; nin Z}\\\frac{2-cos4x+(4-sqrt3), sin2x+2, cos2x-2sqrt3x}{1+2cos2x}-1=0$

$frac{2-cos4x+(4-sqrt3)sin2x+2cos2x-2sqrt3-(1+2cos2x)}{1+2cos2x}=0\\1-cos4x+(4-sqrt3)sin2x-2sqrt3=0; ; ,; ; [; cos4x=1-2sin^22x; ]\\2sin^22x+(4-sqrt3)sin2x-2sqrt3=0\\D=(4-sqrt3)^2+16sqrt3=(16-8sqrt3+3)+16sqrt3=16+8sqrt3+3=(4+sqrt3)^2\\sin2x=frac{-(4-sqrt3)pm sqrt{(4+sqrt3)^2}}{2cdot 2}=frac{-4+sqrt3pm (4+sqrt3)}{4}=left [{{frac{sqrt3}{2}} atop {-2}} right. \\a); ; sin2x=-2<-1; ; to ; ; xin varnothing ; ,t.k.; |sinx|leq 1; ;$

$b); ; sin2x=frac{sqrt3}{2}; ,; ; 2x=left [ {{frac{pi }{3}+2pi m; ,; min Z } atop {frac{2pi }{3}+2pi m,; min z}} right. ; ; ,; ; x=left [ {{frac{pi}{6}+pi m,; min Z} atop {frac{pi}{3}+pi m; notin ODZ}} right.\\Otvet:; ; x=frac{pi }{6}+pi m; ,; min Z; .$