Тип 16 № 14107

Преобразуем уравнение: $$6(1 - \sin^2 x) + 5\sqrt{2}\sin x + 2 = 0$$ $$-6\sin^2 x + 5\sqrt{2}\sin x + 8 = 0$$ $$6\sin^2 x - 5\sqrt{2}\sin x - 8 = 0$$

Сделаем замену: $t = \sin x \in [-1; 1]$: $$6t^2 - 5\sqrt{2}t - 8 = 0$$ $$D = 50 + 4\cdot8\cdot6 = 242$$ $$t = \frac{5\sqrt{2} \pm \sqrt{242}}{12}$$ $$t = \frac{5\sqrt{2} \pm 11\sqrt{2}}{12}$$ $$t = \frac{4\sqrt{2}}{3} \quad \notin [-1; 1] \quad \text{или} \quad t = -\frac{\sqrt{2}}{2}$$ Обратная замена: $$\sin x = -\frac{\sqrt2}{2}$$ $$x = -\frac{\pi}{4} + 2\pi k \quad \text{или} \quad x = -\frac{3\pi}{4} + 2\pi k, k \in Z$$