2cos(2 в квадрате)x-cos x-1=0

Пусть cos x = t , тогда: 2t² - t - 1 = 0  D = [latex]b^{2} - 4ac [/latex] = 1 + 8 = 9 = 3² [latex] t_{1} = \frac{ b^{2} + \sqrt{D} }{2a} = \frac{1 + 3}{4} = 1[/latex] [latex] t_{2} = \frac{ b^{2} - \sqrt{D} }{2a} = \frac{1 - 3}{4} = - \frac{1}{2} [/latex] Вернемся к cosx  1) cosx = 1  [latex] x_{1} = 2pi*n[/latex], n∈Z 2) cosx = -1/2  [latex] x_{2} = +- \frac{4*pi}{6} + 2 pi*k[/latex], k∈Z