Question

|ctg(2x-п/2)|=1/cos^2 2x -1
Help me please

Answer 1

$\Big |ctg(2x-\frac{\pi}{2})\Big |=\frac{1}{cos^22x}-1\\\\\star \; \; 1+tg^2\alpha =\frac{1}{cos^2\alpha }\; \; \; \Rightarrow \; \; \; \frac{1}{cos^2\alpha }-1=tg^2\alpha \; \; \star \\\\\Big |-ctg(\frac{\pi}{2}-2x)\Big |=tg^22x\\\\|-tg2x|=tg^22x\\\\|tg2x|=tg^22x\\\\a)\; \; tg2x\geq 0\; \Rightarrow \; \; \pi n\leq 2x\leq \frac{\pi}{2}+\pi n\; \; ,\; \; \frac{\pi n}{2}\leq x\leq \frac{\pi}{4}+\frac{\pi n}{2}\; \; ,\; n\in Z\; ,\\\\tg2x=tg^22x\; \; ,\; \; tg2x\cdot (tg2x-1)=0\; \; \Rightarrow \; \; tg2x=0\; ,\; \; tg2x=1$

$tg2x=0\; \; \to \; \; 2x=\pi n\; ,\; \; x=\frac{\pi n}{2}\; ,\; n\in Z\\\\tg2x=1\; \; \to \; \; 2x=\frac{\pi}{4}+\pi n\; ,\; \; x=\frac{\pi }{8}+\frac{\pi n}{2}\; ,\; n\in Z\\\\b)\; \; tg2x<0\; \Rightarrow \; \; -\frac{\pi}{2}+\pi n<2x<\pi n\; \; ,\; \; -\frac{\pi}{4}+\frac{\pi n}{2}<x<\frac{\pi n}{2}\; ,\; n\in Z\\\\-tg2x=tg^22x\; ,\; \; tg2x(tg2x+1)=0\; \Rightarrow \; \; tg2x=0\; ,\; \; tg2x=-1\\\\tg2x=0\; \; \to \; \; x=\frac{\pi n}{2}\; ,\; n\in Z\\\\tg2x=-1\; \; \to \; \; 2x=-\frac{\pi}{4}+\pi n\; ,\; \; x=-\frac{\pi}{8}+\frac{\pi n}{2}\; ,\; n\in Z$

$c)\; \; x\in [\; 0,\frac{\pi}{2}\; ]\; :\; \; x=0\; ,\; x=\frac{\pi}{8}\; .$

$Otvet:\; 1)\; \; x=\frac{\pi n}{2}\; \; ,\; \; x=\pm \frac{\pi }{8}+\frac{\pi n}{2}\; ,\; n\in Z\; ,\\\\2)\; \; x=0\; ,\; \; x=\frac{\pi}{8}\; .$