Question

50 баллов. срочно!

2<2^(sinx/1-cosx)^2<8

Answer 1

$2<2^{(\frac{sinx}{1-cosx})^2}<8\; \; ,\; \; \; \; ODZ:\; \; cosx\ne 1\; ,\; \; x\ne 2\pi n\; ,\; n\in Z\\\\\frac{sinx}{1-cosx}=\frac{2sin\frac{x}{2}\cdot cos\frac{x}{2}}{2sin^2\frac{x}{2}}=\frac{cos\frac{x}{2}}{sin\frac{x}{2}}=ctg\frac{x}{2}\\\\2<2^{ctg^2\frac{x}{2}}<2^3\\\\1<ctg^2\frac{x}{2}<3\; \; \Rightarrow \; \; \; \left \{ {ctg^2\frac{x}{2}-3<0} \atop {ctg^2\frac{x}{2}-1>0}} \right. \; \; \left \{ {{(ctg\frac{x}{2}-\sqrt3)(ctg\frac{x}{2}+\sqrt3)<0 } \atop {(ctg\frac{x}{2}-1)(ctg\frac{x}{2}+1)>0}} \right.$

$\left \{ {{-\sqrt3<ctg\frac{x}{2}<\sqrt3} \atop {ctg\frac{x}{2}>1\; ili\; \; ctg\frac{x}{2}<-1}} \right. \\\\a)\; \; -\sqrt3<ctg\frac{x}{2}<\sqrt3\; \; \; \; \Rightarrow\; \; \; \frac{\pi}{6}+\pi n<\frac{x}{2}<\frac{5\pi}{6}+\pi n\; ,\; n\in Z\\\\\frac{\pi}{3}+2\pi n<x<\frac{5\pi }{3}+2\pi n\; ,\; n\in Z\\\\b)\; \; ctg\frac{x}{2}>1\; \; \Rightarrow \; \; \pi k<\frac{x}{2}<\frac{\pi}{4}+\pi k\; ,\; k\in Z\\\\2\pi k<x<\frac{\pi}{2}+2\pi k\; ,\; k\in Z$

$ctg\frac{x}{2}<-1\; \; \Rightarrow \; \; \; \frac{3\pi }{4}+\pi m<\frac{x}{2}<\pi +\pi m\; ,\; m\in Z\\\\\frac{3\pi }{2}+2\pi m<x<2\pi +2\pi m\; ,\; m\in Z\\\\Otvet:\; \; x\in (\frac{\pi }{3}+2\pi n\, ;\,\frac{\pi }{2}+2\pi n)\cup (\frac{3\pi }{2}+2\pi n\, ;\, \frac{5\pi }{3}+2\pi n)\; ,\; n\in Z\; .$