Question

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

Answer 1

$\frac{2cos(x)+sin^2(x)}{ctg(x)-sin(2x)} =tg(2x)\\\frac{2cos(x)+sin^2(x)}{\frac{cos(x)}{sin(x)} -sin(2x)} =\frac{sin(2x)}{cos(2x)} \\\frac{2cos(x)+sin^2(x)}{\frac{cos(x)-sin(x)sin(2x)}{sin(x)} } =\frac{sin(2x)}{cos(2x)} \\\frac{sin(2x)+sin^3(x)}{cos(x)-sin(x)sin(2x)} -\frac{sin(x)}{cos(x)} =0\\\frac{2sin(x)cos(x)+sin^3(x)}{\sqrt{1-sin^2(x)}-sin(x)*2sin(x)cos(x)} -\frac{sin(x)}{\sqrt{1-sin^2(x)}} =0\\ \frac{2sin(x)\sqrt{1-sin^2(x)}+sin^3(x)}{\sqrt{1-sin^2(x)}-sin(x)*2sin(x)\sqrt{1-sin^2(x)}} -\frac{sin(x)}{\sqrt{1-sin^2(x)}} =0\\sin(x)=t,-1\leq t\leq 1\\\frac{2t*\sqrt{1-t^2}+t^3}{\sqrt{1-t^2}-t*2t\sqrt{1-t^2}} -\frac{t}{\sqrt{1-t^2}} =0\\\frac{2t\sqrt{1-t^2}+t^3-t(1-t*2t)}{\sqrt{1-t^2}(1-t*2t)} =0 \\\sqrt{1-t^2} (1-t*2t)\neq 0\\\sqrt{1-t^2}\neq0\\x\neq1\\t\neq-1\\1-2x^2\neq0\\t\neq\frac{\sqrt{2}}{2}\\t\neq-\frac{\sqrt{2}}{2} \\ \sqrt{1-t^2} \geq 0\\-1\leq t\leq 1\\2t\sqrt{1-t^2} =-3t^3+t\\4t^2(1-t^2)=t^2-6t^4+9t^6\\3t^2+2t^4-9t^6=0\\t^2(3+2t^2-9t^4)=0\\t^2=0\\3+2t^2-9t^4=0\\t^2=y\\3+2y-9y^2=0\\9y^2-2y-3=0\\D_1=1+27=28\\y_1=\frac{1+\sqrt{28}}{9} \\y_2=\frac{1-\sqrt{28}}{9} \\t_2=\frac{\sqrt{1+\sqrt{28}}}{3} \\t_3=-\frac{\sqrt{1+\sqrt{28}}}{3} \\\frac{1-\sqrt{28}}{9} =(-0,5) ;5\frac{1-5,5}{9}=(-0,5)\\t_1=0\\t_2=\frac{\sqrt{1+5,5}}{3} =\frac{\sqrt{6,5}}{3} ;2$

0,7 и -0,7 ∉ ОДЗ

t=0\\ [/tex] sin(x)=0\x=\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\\x\neq \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\\x\neq \frac{\pi}{2} +\pi k\\1-2sin^2(x)=0\\cos(2x)\neq 0\\x\neq \frac{\pi}{4} +\frac{\pi k}{2} \\x\neq \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