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

$\frac2cos(x)+sin^2(x)ctg(x)-sin(2x) =tg(2x)\\\frac2cos(x)+sin^2(x)\fraccos(x)sin(x) -sin(2x) =\fracsin(2x)cos(2x) \\\frac2cos(x)+sin^2(x)\fraccos(x)-sin(x)sin(2x)sin(x) =\fracsin(2x)cos(2x) \\\fracsin(2x)+sin^3(x)cos(x)-sin(x)sin(2x) -\fracsin(x)cos(x) =0\\\frac2sin(x)cos(x)+sin^3(x)\sqrt1-sin^2(x)-sin(x)*2sin(x)cos(x) -\fracsin(x)\sqrt1-sin^2(x) =0\\ \frac2sin(x)\sqrt1-sin^2(x)+sin^3(x)\sqrt1-sin^2(x)-sin(x)*2sin(x)\sqrt1-sin^2(x) -\fracsin(x)\sqrt1-sin^2(x) =0\\sin(x)=t,-1\leq t\leq 1\\\frac2t*\sqrt1-t^2+t^3\sqrt1-t^2-t*2t\sqrt1-t^2 -\fract\sqrt1-t^2 =0\\\frac2t\sqrt1-t^2+t^3-t(1-t*2t)\sqrt1-t^2(1-t*2t) =0 \\\sqrt1-t^2 (1-t*2t)\neq 0\\\sqrt1-t^2\neq0\\x\neq1\\t\neq-1\\1-2x^2\neq0\\t\neq\frac\sqrt22\\t\neq-\frac\sqrt22 \\ \sqrt1-t^2 \geq 0\\-1\leq t\leq 1\\2t\sqrt1-t^2 =-3t^3+t\\4t^2(1-t^2)=t^2-6t^4+9t^6\\3t^2+2t^4-9t^6=0\\t^2(3+2t^2-9t^4)=0\\t^2=0\\3+2t^2-9t^4=0\\t^2=y\\3+2y-9y^2=0\\9y^2-2y-3=0\\D_1=1+27=28\\y_1=\frac1+\sqrt289 \\y_2=\frac1-\sqrt289 \\t_2=\frac\sqrt1+\sqrt283 \\t_3=-\frac\sqrt1+\sqrt283 \\\frac1-\sqrt289 =(-0,5) ;5lt;\sqrt28 lt;6 =gt;\sqrt28 =(5,5)=gt;\frac1-5,59=(-0,5)\\t_1=0\\t_2=\frac\sqrt1+5,53 =\frac\sqrt6,53 ;2lt;\sqrt6,5lt;3 =gt;\sqrt6,5 =2,1\\ \frac2,13=0,7\\-\frac\sqrt1+5,53 =-0,7 \\ODZ:\\-1lt;tlt;-\frac\sqrt22;-\frac\sqrt22 lt;tlt;\frac\sqrt22 ;\frac\sqrt22 lt;tlt;1$

0,7 и -0,7 ОДЗ

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

kZ

[/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\\x\neq \frac\pi4 +\frac\pi k2 \\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\\x\neq \frac\pi2 +\pi k\\1-2sin^2(x)=0\\cos(2x)\neq 0\\x\neq \frac\pi4 +\frac\pi k2 \\x\neq \left \ \frac\pi4+\frac\pi k2 \atop \frac\pi2 +\pi k \right. [/tex]

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

Ответ:x=k,kZ

О проекте

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

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