Question

Задание Решите уравнение:

Answer 1

$\displaystyle\bf\\Cos\Big(2x-\frac{\pi }{3} \Big)=\frac{\sqrt{3} }{2} \\\\\\2x-\frac{\pi }{3}=\pm \ arcCos\frac{\sqrt{3} }{2} +2\pi n,n\in Z\\\\\\2x-\frac{\pi }{3} =\pm \frac{\pi }{6}+2\pi n,n\in Z\\\\\\2x=\pm \ \frac{\pi }{6} +\frac{\pi }{3} +2\pi n,n\in Z\\\\\\\boxed{x=\pm \ \frac{\pi }{12} +\frac{\pi }{6} +\pi n,n\in Z}\\\\\\1)\\\\x_{1} =\frac{\pi }{12} +\frac{\pi }{6} +\pi n,n\in Z\\\\\\x_{1} =\frac{\pi }{4} +\pi n,n\in Z\\\\\\2)$

$\displaystyle\bf\\x_{2} =-\frac{\pi }{12} +\frac{\pi }{6} +\pi n,n\in Z\\\\\\x_{2} =\frac{\pi }{12} +\pi n,n\in Z$