Question

Помогите пожалуйста решить задачу ​

Answer 1

Ответ:

Правило:     $\bf |x|\geq a\ \ \Rightarrow \ \ \left[\begin{array}{l}\bf x\geq a\\\bf x\leq -a\end{array}\right$   .  

Заметьте, это не система неравенств, а совокупность неравенств .

   $|\, tg8x\, |\geq \sqrt3\ \ \ \Rightarrow \ \ \ \left[\begin{array}{l}tg8x\geq \sqrt3\\tg8x\leq -\sqrt3\end{array}\right$

$\left[\begin{array}{l}\dfrac{\pi}{3}+\pi n\leq 8x < \dfrac{\pi}{2}+\pi n\\-\dfrac{\pi}{2}+\pi k < 8x\leq -\dfrac{\pi}{3}+\pi k\end{array}\right\ \ \ \ \left[\begin{array}{l}\dfrac{\pi}{24}+\dfrac{\pi n}{8}\leq x < \dfrac{\pi}{16}+\dfrac{\pi n}{8}\\-\dfrac{\pi}{16}+\dfrac{\pi k}{8} < x\leq -\dfrac{\pi}{24}+\dfrac{\pi k}{8}\end{array}\right\ \ ,\ \ n,k\in Z$

$Otvet:\ x\in \Big(-\dfrac{\pi}{16}+\dfrac{\pi n}{8}\ ;\ -\dfrac{\pi}{24}+\dfrac{\pi n}{8}\ \Big]\cup \Big[\ \dfrac{\pi}{24}+\dfrac{\pi n}{8}\ ;\ \dfrac{\pi}{16}+\dfrac{\pi n}{8}\ \Big)\ ,\ \ n\in Z$   .

Если условие было другим, и в аргументе 3х, а не 8х , то аналогично

   $|\, tg3x\, |\geq \sqrt3\ \ \ \Rightarrow \ \ \ \left[\begin{array}{l}tg3x\geq \sqrt3\\tg3x\leq -\sqrt3\end{array}\right$

$\left[\begin{array}{l}\dfrac{\pi}{3}+\pi n\leq 3x < \dfrac{\pi}{2}+\pi n\\-\dfrac{\pi}{2}+\pi k < 3x\leq -\dfrac{\pi}{3}+\pi k\end{array}\right\ \ \ \ \left[\begin{array}{l}\dfrac{\pi}{9}+\dfrac{\pi n}{3}\leq x < \dfrac{\pi}{6}+\dfrac{\pi n}{3}\\-\dfrac{\pi}{6}+\dfrac{\pi k}{3} < x\leq -\dfrac{\pi}{9}+\dfrac{\pi k}{3}\end{array}\right\ \ ,\ \ n,k\in Z$

$Otvet:\ x\in \Big(-\dfrac{\pi}{6}+\dfrac{\pi n}{3}\ ;\ -\dfrac{\pi}{9}+\dfrac{\pi n}{3}\ \Big]\cup \Big[\ \dfrac{\pi}{9}+\dfrac{\pi n}{3}\ ;\ \dfrac{\pi}{6}+\dfrac{\pi n}{3}\ \Big)\ ,n\in Z$   .