Question

решите уравнение 1/sin^2x+tgx-ctgx=4​

Answer 1

$\dfrac{1}{\sin^{2}\left(2\,x\right)}+\mathrm{tg}\left(x\right)-\mathrm{ctg}\left(x\right)-4=0\Leftrightarrow \\\Leftrightarrow \dfrac{\left(\sin^{2}\left(x\right)-4\,\cos\left(x\right)\,\sin\left(x\right)-\cos^{2}\left(x\right)\right)\,\sin^{2}\left(2\,x\right)+\cos\left(x\right)\,\sin\left(x\right)}{\cos\left(x\right)\,\sin\left(x\right)\,\sin^{2}\left(2\,x\right)}=0\overset{\cos\left(x\right)\,\sin\left(x\right)\,\sin^{2}\left(2\,x\right)\neq 0}{\Leftrightarrow }$$\Leftrightarrow\left(\sin^{2}\left(x\right)-4\,\cos\left(x\right)\,\sin\left(x\right)-\cos^{2}\left(x\right)\right)\,\sin^{2}\left(2\,x\right)+\cos\left(x\right)\,\sin\left(x\right)=0 \Leftrightarrow$$\sin\left(x\right)=\dfrac{2\,\mathrm{tg}\left(\frac{\mathrm{x}}{2}\right)}{1+\mathrm{tg}^2\left(\frac{\mathrm{x}}{2}\right)},\cos\left(\mathrm{x}\right)=\dfrac{1-\mathrm{tg}^2\left(\frac{\mathrm{x}}{2}\right)}{1+\mathrm{tg}^2\left(\frac{\mathrm{x}}{2}\right)},\mathrm{tg}\frac{x}{2}=u\Rightarrow$$\Rightarrow -\dfrac{8\,u\,\left(1-u^{2}\right)^{3}}{\left(u^{2}+1\right)^{4}}-\dfrac{64\,u^{2}\,\left(1-u^{2}\right)^{2}}{\left(u^{2}+1\right)^{4}}+\dfrac{32\,u^{3}\,\left(1-u^{2}\right)}{\left(u^{2}+1\right)^{4}}+1=0$$\overset{\left(u^{2}+1\right)^{4}\neq 0}{\Rightarrow }u^{8}+8\,u^{7}-60\,u^{6}-56\,u^{5}+134\,u^{4}+56\,u^{3}-60\,u^{2}-8\,u+1=0\\u^4+\frac{1}{u^4}+8\left ( u^3+\frac{1}{u^3} \right )-60\left ( u^2+\frac{1}{u^2} \right )-56\left ( u+\frac{1}{u} \right )+134=0\\u+\frac{1}{u}=t\Rightarrow t^4-56t^2+16+8t^3-32t=0\Leftrightarrow t^2+\frac{16}{t^2}-56+8\left ( t-\frac{4}{t} \right )=0$$t-\frac{4}{t}=z\Rightarrow z^2+8z-48=0\Rightarrow z=\left \{ -12,4 \right \}\Rightarrow t=\left \{ -6\pm 2\sqrt{10},2\pm 2\sqrt{2} \right \}$$u_{1,2,3,4}=\left \{ 1+\sqrt{2}\pm \frac{\sqrt{8\sqrt{2}+16}}{2},1-\sqrt{2}\pm \frac{\sqrt{16-8\sqrt{2}}}{2} \right \}\\u_{5,6,7,8}=\left \{ -3+\sqrt{10}\pm \sqrt{20-6\sqrt{10}},-3-\sqrt{10}\pm \sqrt{6\sqrt{10}+20} \right \}$

$\begin{gathered}x_{1}=2\,\mathrm{ctg}\left(\frac{\sqrt{8\,\sqrt{2}+16}}{2}+\sqrt{2}+1\right)+2\,\pi\,\mathrm{k}\\x_{2}=2\,\pi\,\mathrm{k}-2\,\mathrm{ctg}\left(\frac{\sqrt{8\,\sqrt{2}+16}}{2}-\sqrt{2}-1\right)\\x_{3}=2\,\mathrm{ctg}\left(\frac{\sqrt{16-8\,\sqrt{2}}}{2}-\sqrt{2}+1\right)+2\,\pi\,\mathrm{k}\\x_{4}=2\,\pi\,\mathrm{k}-2\,\mathrm{ctg}\left(\frac{\sqrt{16-8\,\sqrt{2}}}{2}+\sqrt{2}-1\right)\end{gathered}$$\begin{gathered}x_{5}=2\,\mathrm{ctg}\left(\sqrt{10}+\sqrt{20-6\,\sqrt{10}}-3\right)+2\,\pi\,\mathrm{k}\\x_{6}=2\,\mathrm{ctg}\left(\sqrt{10}-\sqrt{20-6\,\sqrt{10}}-3\right)+2\,\pi\,\mathrm{k}\\x_{7}=2\,\mathrm{ctg}\left(\sqrt{6\,\sqrt{10}+20}-\sqrt{10}-3\right)+2\,\pi\,\mathrm{k}\\x_{8}=2\,\pi\,\mathrm{k}-2\,\mathrm{ctg}\left(\sqrt{6\,\sqrt{10}+20}+\sqrt{10}+3\right)\end{gathered}$