Question

Sin²x-4sin x×cos x+3 cos² x=0

Answer 1

Ответ:

$x_1 = \frac{\pi}{4} + \pi k;\ \ \ x_2= \arctan (\frac{1}{3} ) + \pi k, k \in \mathbb {Z}$

Пошаговое объяснение:

$\sin ^2 x - 4\sin x \cos x + 3 \cos^2 x = 0 \ | : \ \sin^2 x \neq 0; \ x \neq \pi k, k \in \mathbb {Z}\\ 1 - 4\tan x + 3\tan^2 x = 0\\3\tan^2 x - 4\tan x + 1 = 0\ | \  t = \tan x\\3t^2-4t+1=0\\t_1=1 \ ; t_2 = \frac{1}{3} \\\tan x = 1 \ \ \ \ \tan x = \frac{1}{3}\\x_1 = \frac{\pi}{4} + \pi k;\ \ \ x_2= \arctan (\frac{1}{3} ) + \pi k, k \in \mathbb {Z}$