Question

2sin^2x-(2*корень из 2)*cosx+1=0

Answer 1

Ответ:

$\displaystyle x = \pm \frac{\pi}{4} + 2\pi n, n \in \mathbb{Z}$

Объяснение:

$\displaystyle 2\sin^2x - 2\sqrt 2\cdot \cos x + 1 = 0 ~~~~ \sin^2 x= 1 - \cos^2x \\ \\ 2(1-\cos^2x)-2\sqrt 2 \cos x + 1 = 0 \\ \\ 2 - 2 \cos^2 x - 2\sqrt 2 \cos x + 1 = 0 \\ \\ -2\cos^2 x - 2\sqrt 2 \cos x + 3 = 0$

Замена $\cos x = t,~ -1 \leq t \leq 1$

$\displaystyle -2t^2 - 2\sqrt 2\cdot t + 3 = 0\\ \\ D = 8 + 4\cdot 2 \cdot 3 = 32 \\ \\ x_1 = \frac{2\sqrt 2 + 4\sqrt 2}{-4} = -\frac{3\sqrt 2}{2} < -1\\ \\ x_2 = \frac{2\sqrt 2 - 4\sqrt 2}{-4} = \frac {\sqrt 2} {2}$

$\displaystyle \cos x = \frac{\sqrt 2}{2} \\ \\ x = \pm \text{arccos} \frac{\sqrt 2}{2} + 2\pi n, n \in \mathbb{Z} \\ \\ x = \pm \frac{\pi}{4} + 2\pi n, n \in \mathbb{Z}$