Question

Решить уравнение
2cos(п/3-3x)+√3=0

Answer 1

Ответ:

$\displaystyle 2cos\Big(\frac{\pi }{3}-3x\Big)+\sqrt3=0\\\\\\cos\Big(\frac{\pi }{3}-3x\Big)=-\frac{\sqrt3}{2}\\\\\\\frac{\pi }{3}-3x=\pm arccos\Big(-\frac{\sqrt3}{2}\Big)+2\pi n\ ,\ n\in Z\\\\\\\frac{\pi }{3}-3x=\pm \Big(\pi -arccos\frac{\sqrt3}{2}\Big)+2\pi n\ ,\ n\in Z\\\\\\\frac{\pi }{3}-3x=\pm \Big(\pi -\frac{\pi }{6}\Big)+2\pi n\ ,\ n\in Z$

$\displaystyle \frac{\pi }{3}-3x=\pm \frac{5\pi}{6}+2\pi n\ ,\ n\in Z\\\\\\3x=\frac{\pi }{3}\pm \frac{5\pi}{6}-2\pi n\ ,\ n\in Z\\\\\\x=\frac{\pi }{9}\pm \frac{5\pi}{18}-\frac{2\pi n}{3}\ ,\ n\in Z\\\\\\x=\left[\begin{array}{l}\dfrac{7\pi}{18}-\dfrac{2\pi n}{3}\ ,\ n\in Z\\\ -\dfrac{\pi}{6}-\dfrac{2\pi n}{3}\ ,\ n\in Z\end{array}\right$