Question

2cos^2(5x)-3sin5x*cos5x+sin^2(5x)=0 решите

Answer 1

$2\cos^2(5x)-3\sin5x*\cos5x+\sin^2(5x)=0\\ 2-3\tan 5x+\tan^2 5x=0\\ (\tan 5x - 1)(\tan 5x -2)=0\\ \tan 5x_1 = 1; \tan 5x_2 = 2\\ \begin{cases} 5x_1 = \frac{\pi}{4} + \pi k\\ 5x_2 = \arctan 2 \end{cases}\\ \begin{cases} x_1 = \frac{\pi+4\pi k}{20}\\ x_2 = \frac{\arctan 2}{5} \end{cases}$