Question

Решите уравнение: cos2x−√2cos(3π/2+x)−1=0

Answer 1

$\cos(2x) - \sqrt{2} \cos( \frac{3\pi}{2} + x) - 1 = 0 \\ \cos(2x) - \sqrt{2} \sin(x) - 1 = 0 \\ 1 - 2 \sin ^{2} (x) - \sqrt{2 } \sin(x) - 1 = 0 \\ - 2 \sin^{2} (x) - \sqrt{2} \sin(x) = 0 \\ - \sin(x) (2 \sin(x) + \sqrt{2} ) = 0 \\ \sin(x) = 0 \: \: \: and \: \: \sin(x) = - \frac{ \sqrt{2} }{2} \\ x = \pi \: k \: \: \: and \: \: \: \: \: \: \: \: x = - \frac{\pi}{4} + 2\pi \: k \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x = \pi + \frac{\pi}{4} = \frac{5\pi}{4} + 2\pi \: k$
$k\in \mathbb z$