Question

Решить уравнение:
2sin^2x - 2sin2x +1 = 0

Answer 1

$2 \sin^{2} (x) - 2 \sin(2x) + 1 = 0 \\ 2 \times \frac{1 - \cos(2x) }{2} - 2 \sin(2x) + 1 = 0 \\ - \cos(2x) - 2 \sin(2x) = 0$

Разделим уравнение на (-cos(2x)) :

$1 + 2\tan(2x) = 0 \\ \tan(2x) = - \frac{1}{2} \\ 2x = - arctg \frac{1}{2} + \pi \times k \\ x = - \frac{1}{2} arctg \frac{1}{2} + \pi \times k$

Answer 2

$2sin^2x-2sin2x+1=0\\2sin^2x-4sinxcosx+sin^2x+cos^2x=0\\3sin^2x-4sinxcosx+cos^2x=0\\\left.\begin{matrix}3tg^2x-4tgx+1=0\end{matrix}\right|x\neq \frac{\pi}{2}+\pi k,k\in \mathbb{Z}\\tgx=\left \{ 1;\frac{1}{3} \right \}\Rightarrow x=\left \{ \frac{\pi}{4}+\pi k;arctg\frac{1}{3}+\pi k \right \},k\in \mathbb{Z}$