Помогите решить уравнение 1.sin 2x=-√2/2; 2. 2cos²x+cos x-6=0

1)
$\sin(2x ) = - \sqrt{2} \div 2 \\ 2x = - {1}^{k} \times arcsin ( - \frac{ \sqrt{2} }{2}) + \pi \times k \\ 2x = - {1}^{k} \times ( - \frac{\pi}{4}) + \pi \times k \\ x = - {1}^{k} \times ( - \frac{\pi}{8} ) + \frac{\pi \times k}{2}$
2)
$2 {cos}^{2} x + \cos(x ) - 6 = 0 \\ \cos(x) = t \\2 {t}^{2} + t - 6 = 0 \\ d = 1 + 24 = 25 \\ x1 = \frac{ - 1 + 5}{4} = 1 \\ x2 = \frac{ - 1 - 5}{4} = - \frac{6}{4} \\ \cos(x) = 1 \\ x = 2\pi \: k$