Question

Решите уравнение cos( П/6-2x)=-1/2
Распишите как решать, пж
Варианты ответа:
1)-П/3+Пк,к€z
2)-П/12+-П/3+Пк, к€z
3)П/6+-2П/3+Пк, к€z
4)(-1)^k П/6+2Пк, к€z

Answer 1

$cos\Big(\dfrac{\pi}{6}-2x\Big)=-\dfrac{1}{2}\\\\\\\dfrac{\pi}{6}-2x=\pm arccos\Big(-\dfrac{1}{2}\Big)+2\pi n\ \ ,\ \ n\in Z\\\\\\\dfrac{\pi}{6}-2x=\pm \Big(\pi -arccos\dfrac{1}{2}\Big)+2\pi n\ \ ,\ \ n\in Z\\\\\\\dfrac{\pi}{6}-2x=\pm \Big(\pi -\dfrac{\pi}{3}\Big)+2\pi n\ \ ,\ \ n\in Z\\\\\\\dfrac{\pi}{6}-2x=\pm \dfrac{2\pi}{3}+2\pi n\ \ ,\ \ n\in Z\\\\\\2x=\dfrac{\pi}{6}\mp \dfrac{2\pi}{3}-2\pi n\ ,\ n\in Z$

$x=\dfrac{\pi}{12}\mp \dfrac{\pi}{3}-\pi n\ ,\ n\in Z\ \ \ ili\ \ \ \ \boxed {\ x=\dfrac{\pi}{12}\pm \dfrac{\pi}{3}+\pi n\ ,\ n\in Z\ }$

$ili\ \ \ \ \ x=\left[\begin{array}{l}-\dfrac{3\pi}{12}+\pi n\ ,\ n\in Z\\\\\ \ \dfrac{5\pi}{12}+\pi n\ ,\ n\in Z\end{array}\right\ \ \ \ ili\ \ \ \ \ x=\left[\begin{array}{l}-\dfrac{\pi}{4}+\pi n\ ,\ n\in Z\\\\\ \ \dfrac{5\pi}{12}+\pi n\ ,\ n\in Z\end{array}\right$