Question

3sin^2(x)-4sin(x)cos(x)+cos^2(x)=0

Answer 1

$3 {sin}^{2} x - 4sinxcosx + {cos}^{2} x = 0 \\ \frac{3 {sin}^{2} x}{ {cos}^{2} x} - \frac{4sinxcosx}{ {cos}^{2}x } + \frac{ {cos}^{2} x}{ {cos}^{2}x } = 0 \\ 3 {tg}^{2} x - 4tgx + 1 = 0 \\ tgx = t \\ 3 {t}^{2} - 4t + 1 = 0 \\ d = {b}^{2} - 4ac = 16 - 4 \times 3 = 16 - 12 = 4 \\ t1 = \frac{4 + 2}{2 \times 3} = \frac{6}{6} = 1 \\ t2 = \frac{ 4 - 2}{2 \times 3} = \frac{2}{6} = \frac{1}{3} \\ 1)tgx = 1 \\ x = \frac{\pi}{4} + \pi n \\ 2)tgx = \frac{1}{3} \\ x = arctg \frac{1}{3} + \pi n$
Ответ: p/4 + pi*n, n € Z; arctg1/3 + pi*n, n € Z.