Вопрос в картинках...

Решите задачу:

$6sinx - 4cosx = \frac{1}{sin x}|*sin x\\ 6sin^2x - 4sinxcosx - 1 = 0\\ 6sin^2x - 4sinxcosx - sin^2x - cos^2x = 0\\ 5sin^2x - 4sinxcosx - cos^2x = 0 |*\frac{1}{cos^x}\\ 5tg^2x - 4tgx - 1 = 0\\ t = tg x\\ 5t^2 - 4t - 1=0\\ D_1 = 4 + 5 = 9\\ t_1 = -\frac{1}{5}, t_2 = 1\\ tg x = 1\\ x = \frac{\pi}{4} + \pi n\\ tgx = -\frac{1}{5}\\ x = -arctg{1}{5} + \pi k$