Question

2cos(2 в квадрате)x-cos x-1=0

Answer 1

Пусть cos x = t , тогда:

2t² - t - 1 = 0 
D = $b^{2} - 4ac$ = 1 + 8 = 9 = 3²
$t_{1} = frac{ b^{2} + sqrt{D} }{2a} = frac{1 + 3}{4} = 1$
$t_{2} = frac{ b^{2} - sqrt{D} }{2a} = frac{1 - 3}{4} = - frac{1}{2}$

Вернемся к cosx 

1) cosx = 1 
$x_{1} = 2pi*n$, n∈Z

2) cosx = -1/2 
$x_{2} = +- frac{4*pi}{6} + 2 pi*k$, k∈Z