ЕГЭ по профильной математике 13 задание 4cos^2x+2(√2-1)sin(π/2-x)-√2=0 помогите решить

Решите задачу:

$4 \cos^{2} (x) + 2( \sqrt{2} - 1) \sin( \frac{\pi}{2} - x) - \sqrt{2} = 0 \\ 4 \cos^{2} (x) + 2( \sqrt{2} - 1) \cos(x) - \sqrt{2} = 0 \\ \cos(x) = t \\ 4 {t}^{2} + 2( \sqrt{2} - 1)t - \sqrt{2} = 0 \\ D = (2 (\sqrt{2} - 1)) ^{2} + 4 \times 4 \sqrt{2} = 4(2 - 2 \sqrt{2} + 1) + \\ + 16 \sqrt{2} = 8 - 8 \sqrt{2} + 4 + 16 \sqrt{2} = 8 + 8 \sqrt{2} + 4 = \\ = (2 + 2 \sqrt{2} )^{2} \\ t_{1} = \frac{ - 2\sqrt{2} + 2 - 2 - 2 \sqrt{2} }{8} = - \frac{ \sqrt{2} }{2} \\ t_{2} = \frac{ - 2 \sqrt{2} + 2 + 2 + 2 \sqrt{2} }{8} = \frac{1}{2} \\$

$1) \cos(x) = - \frac{ \sqrt{2} }{2} \\ x = \frac{3\pi}{4} + 2\pi n \\ x = \frac{5\pi}{4} + \pi n, \: n \in \mathbb Z$

$2) \cos(x) = \frac{1}{2} \\ x = ± \frac{\pi}{3} + 2\pi m, \: m \in \mathbb Z$