Помогите пожалуйста срочно 3cos^2x-sin^2x+2sinx cosx=0

$3 \cos^{2}x - \sin^{2}x + 2\sin x \cos x = 0 \ \ \ |:\sin^{2}x \neq 0$

$3\dfrac{\cos^{2}x}{\sin^{2}x} - \dfrac{\sin^{2}x}{\sin^{2}x} + 2\dfrac{\sin x \cos x}{\sin^{2}x} = 0$

$3\, \text{ctg}^{2} \, x - 1 + 2\, \text{ctg} \, x = 0$

Замена: $\text{ctg} \, x = t$

$3t^{2} + 2t - 1 = 0$

$D = 2^{2} - 4 \cdot 3 \cdot (-1) = 16$

$t_{1,2} = \dfrac{-2 \pm \sqrt{16}}{2 \cdot 3} =\dfrac{-2 \pm 4}{6} = \left[\begin{array}{ccc}t_{1} = -1 \\t_{2} = \dfrac{1}{3} \ \, \\\end{array}\right$

Обратная замена:

$1) \ \text{ctg} \, x = -1$

$x = \text{arcctg} \, (-1) + \pi n, \ n \in Z$

$x = \pi - \text{arcctg} \, 1 + \pi n, \ n \in Z$

$x = \pi - \dfrac{\pi}{4} + \pi n, \ n \in Z$

$x = \dfrac{3\pi}{4} + \pi n, \ n \in Z$

$2) \ \text{ctg} \, x = \dfrac{1}{3}$

$x = \text{arcctg} \, \dfrac{1}{3} + \pi n, \ n \in Z$

Ответ: $x = \dfrac{3\pi}{4} + \pi n, \ x = \text{arcctg} \, \dfrac{1}{3} + \pi n, \ n \in Z$