Cos(x)^2+cos(2x)^2=1 СРОЧНО!!

Ответ:

$x = \frac{+}{} \frac{\pi }{6} + \pi k \\$

$x = \frac{\pi }{2} + \pi k$

Пошаговое объяснение:

$cos^{2}x + cos^{2}(2x) = 1\\\\cos^{2}x + cos^{2}(2x) - 1 = 0 \\\\2cos^{2}x + 2cos^{2}(2x) - 2 = 0 \\\\(2cos^{2}x - 1) + 2cos^{2}(2x) - 1 = 0 \\\\cos2x + 2cos^{2}(2x) - 1 = 0\\\\ 2cos^{2}(2x) + cos2x - 1 = 0\\\\$

Пусть $cos2x = t$ , тогда:

$2t^{2} + t - 1 = 0\\\\D = 1^{2} - 4*1*(-1) = 9\\\\x_{1} = \frac{-1 + 3}{2*2} = \frac{2}{4} = \frac{1}{2} \\\\x_{2} = \frac{-1 - 3}{2*2} = \frac{-4}{4} = -1 \\\\$

Решаем 2 уравнения:

$cos2x = \frac{1}{2} \\\\2x = \frac{+}{} \frac{\pi }{3} + 2\pi k \\\\x = \frac{+}{} \frac{\pi }{6} + \pi k \\$

$cos2x = -1\\\\2x = \pi +2\pi k\\\\x = \frac{\pi }{2} + \pi k$