Question

6cos^2x + sin^2x - 5sin x * cos x = 0

Answer 1

$6 \cos {}^{2} (x) + \sin {}^{2} (x) - 5 \sin(x) \cos(x) = 0 \\ \sin {}^{2} (x) - 5 \sin(x) \cos(x) + 6\cos {}^{2} (x) = 0 \\ | \div \cos {}^{2} (x) \ne0 \\ \\ {tg}^{2} x - 5tgx + 6 = 0 \\ \\ tgx = t \\ \\ t {}^{2} - 5t + 6 = 0\\ D = 25 - 24 = 1\\ t_1 = \frac{5 + 1}{2} = 3 \\ t_2 = 2 \\ \\ tgx = 3 \\ x_1 = arctg(3) + \pi \: n \\ \\ tgx = 2 \\ x_2 = arctg(2) + \pi \: n \\ \\ n\in \:Z$