Question

((2cosx+sin^2(x))/(ctgx-sin2x))=tg2x

Answer 1

$frac{2cos(x)+sin^2(x)}{ctg(x)-sin(2x)} =tg(2x)\frac{2cos(x)+sin^2(x)}{frac{cos(x)}{sin(x)} -sin(2x)} =frac{sin(2x)}{cos(2x)} \frac{2cos(x)+sin^2(x)}{frac{cos(x)-sin(x)sin(2x)}{sin(x)} } =frac{sin(2x)}{cos(2x)} \frac{sin(2x)+sin^3(x)}{cos(x)-sin(x)sin(2x)} -frac{sin(x)}{cos(x)} =0\frac{2sin(x)cos(x)+sin^3(x)}{sqrt{1-sin^2(x)}-sin(x)*2sin(x)cos(x)} -frac{sin(x)}{sqrt{1-sin^2(x)}} =0\ frac{2sin(x)sqrt{1-sin^2(x)}+sin^3(x)}{sqrt{1-sin^2(x)}-sin(x)*2sin(x)sqrt{1-sin^2(x)}} -frac{sin(x)}{sqrt{1-sin^2(x)}} =0\sin(x)=t,-1leq tleq 1\frac{2t*sqrt{1-t^2}+t^3}{sqrt{1-t^2}-t*2tsqrt{1-t^2}} -frac{t}{sqrt{1-t^2}} =0\frac{2tsqrt{1-t^2}+t^3-t(1-t*2t)}{sqrt{1-t^2}(1-t*2t)} =0 \sqrt{1-t^2} (1-t*2t)neq 0\sqrt{1-t^2}neq0\xneq1\tneq-1\1-2x^2neq0\tneqfrac{sqrt{2}}{2}\tneq-frac{sqrt{2}}{2} \ sqrt{1-t^2} geq 0\-1leq tleq 1\2tsqrt{1-t^2} =-3t^3+t\4t^2(1-t^2)=t^2-6t^4+9t^6\3t^2+2t^4-9t^6=0\t^2(3+2t^2-9t^4)=0\t^2=0\3+2t^2-9t^4=0\t^2=y\3+2y-9y^2=0\9y^2-2y-3=0\D_1=1+27=28\y_1=frac{1+sqrt{28}}{9} \y_2=frac{1-sqrt{28}}{9} \t_2=frac{sqrt{1+sqrt{28}}}{3} \t_3=-frac{sqrt{1+sqrt{28}}}{3} \frac{1-sqrt{28}}{9} =(-0,5) ;5<sqrt{28} <6 =>sqrt{28} =(5,5)=>frac{1-5,5}{9}=(-0,5)\t_1=0\t_2=frac{sqrt{1+5,5}}{3} =frac{sqrt{6,5}}{3} ;2<sqrt{6,5}<3 =>sqrt{6,5} =2,1\ frac{2,1}{3}=0,7\-frac{sqrt{1+5,5}}{3} =-0,7 \ODZ:\-1<t<-frac{sqrt{2}}{2};-frac{sqrt{2}}{2} <t<frac{sqrt{2}}{2} ;frac{sqrt{2}}{2 <t<1$

0,7 и -0,7 ∉ ОДЗ

t=0\ [/tex] sin(x)=0x=pi k [/tex]

k∈Z

[/tex] ODZ:cos(x)cos(2x)-sin(x)sin(2x)cos(2x)neq 0\cos(2x)(cos(x)-sin(x)sin(2x))neq 0\cos(2x)neq 0\xneq frac{pi}{4} +frac{pi k}{2} \cos(x)-sin(x)sin(2x)neq 0\cos(x)-2sin^2(x)cos(x)neq 0\cos(x)(1-2sin^2(x))neq =0\cos(x)neq 0\xneq frac{pi}{2} +pi k\1-2sin^2(x)=0\cos(2x)neq 0\xneq frac{pi}{4} +frac{pi k}{2} \xneq left { {{frac{pi}{4}+frac{pi k}{2} } atop {frac{pi}{2} }+pi k} right. [/tex]

Первое ОДЗ было сделано на t .Второе ОДЗ было сделано на x

Ответ:x=πk,k∈Z