Question

Cei care știți va rog rapid

Answer 1

$b)\quad \mathrm{tg}\,2x = \mathrm{tg}\,x \\ \\ 2x = k\pi +x \Rightarrow x = k\pi,\quad k\in \mathbb{Z}\\ \\ \\\\ d)\quad \cos\Big(2x+\dfrac{\pi}{2}\Big) = \cos\Big(x-\dfrac{\pi}{2}\Big) \\ \\ \\ 2x+\dfrac{\pi}{2} = \pm\Big(x-\dfrac{\pi}{2}\Big)+2k\pi$

$\Rightarrow \left|\begin{array}{lcl}2 x+\dfrac{\pi}{2}=x-\dfrac{\pi}{2}+2k\pi\\\\2x+\dfrac{\pi}{2} = -x+\dfrac{\pi}{2}+2k\pi\ \end{array}\right. \Rightarrow \left|\begin{array}{lcl}x=-\pi+2k\pi\\\\3x =2k\pi\ \end{array} \right. \Rightarrow \\ \\ \\ \Rightarrow S= \Bigg\{(2k-1)\pi;\,\dfrac{2k\pi}{3}\bigg|\,k\in \mathbb{Z}\Bigg\}$