Question

решите пожалуйста неравенство:корень из 2 TGx CTGx < 2SINx

Answer 1

$\sqrt{2}tgx\cdot ctgx<2\sin x;\\ D(f): \left \{ {{x\neq\pi n} \atop {x\neq\frac\pi2+\pi k}} \right. n,k\in Z;\ \ ==>x\neq\frac{\pi n}{2},\ n\in Z;\\ D(f):x\in\left(\frac{\pi n}{2};\frac\pi2+\frac{\pi n}{2}\right),\ n\in Z;\\ \sqrt2<2\sin x;\\ \frac{\sqrt2}{2}<\sin x;\\ \sin x=\frac{\sqrt2}{2};\\ x=\left(-1\right)^n\cdot\frac\pi4+\pi n= \left \{ {{x=\frac\pi4+2\pi l} \atop {x= -\frac{\pi}{4}+\pi+2\pi m}} \right.==>\\ ==> \left \{ {{x=\frac\pi4+2\pi n} \atop {x= \frac{3\pi}{4}+2\pi m}} \right. \ n,m\in Z;$
$\left \{ {{\frac\pi4+2\pi l<x<\frac{3\pi}{4}+2\pi;\ l\in Z;} \atop {x\neq\frac{\pi m}{2},\ m\in Z;}} \right. \\\frac\pi4+2\pi n<x<\frac\pi2+2\pi n\cup\frac\pi2+2\pi n<x<\frac{3\pi }{4}+2\pi n;\ n\in Z;\\x\in\left(\frac\pi4+2\pi n;\frac\pi2+2\pi n\right)\cup\left(\frac\pi2+2\pi n;\frac{3\pi}{4}+2\pi n\right);n\in Z.$