решите уравнение 1. 2cosx-корень из 2=02 .5cos(2x+П/6)+1=03. sin(x/2-П/5)=15....

Решите задачу:

$2cosx - \sqrt{2} = 0 \\ \\ cosx = \dfrac{ \sqrt{2} }{2} \\ \\ \boxed{x = \pm \dfrac{\pi} {4} + 2 \pi n, \ n \in Z}$

$5cos(2x + \dfrac{\pi }{6} ) + 1 = 0 \\ \\ cos(2x + \dfrac{ \pi }{6} ) = -1 \\ \\ 2x + \dfrac{ \pi }{6} = \pm arccos(- \dfrac{1}{5} ) + 2 \pi n, \ n \in Z \\ \\ \boxed{ x = \pm \dfrac{1}{2} arccos(- \dfrac{1}{5} ) - \dfrac{ \pi }{12} + \pi n, \ n \in Z}$

$sin( \dfrac{x}{2} - \dfrac{ \pi }{5} ) = 1 \\ \\ \dfrac{x}{2} - \dfrac{ \pi }{5} = \dfrac{ \pi }{2} + 2 \pi n, \ n \in Z \\ \\ \boxed{x = \pi + \dfrac{2 \pi }{5} + 4 \pi n, \ n \in Z}$

$sin(2 \pi - x) - cos( \dfrac{3 \pi }{2} + x) + 1 = 0 \\ \\ sin(-x) - cos( \pi + \dfrac{ \pi }{2} + x) = -1 \\ \\ -sinx + cos( \dfrac{ \pi }{2} +x) = -1 \\ \\ -sinx - sinx = -1 \\ \\ -2sinx = -1 \\ \\ sinx = \dfrac{1}{2} \\ \\ \boxed{ x = (-1)^{n} \dfrac{\pi }{6} + \pi n, \ n \in Z}$