Решите уравнения: 1. 2cos x - sqrt(2) = 0; 2. sin(x/2 - pi/6) = 1; 3. cos^2 x + 2sin x + 2 = 0; 4. 6sin^2 x = 5sin x cos x - cos^2 x; 5. 5sin^2 x + 2sin x cos x - cos^2 x = 1; 6. Найдите корни уравнения sin 3x + cos 3x = 0, принадлежащие отрезку [0, 6].

1. $2\cos x - \sqrt{2} = 0 \Rightarrow \cos x = \frac{\sqrt{2}}{2} \Rightarrow x = \pm \frac{\pi}{4} + 2\pi n, n \in \mathbb{Z}.$ 2. $\sin\left(\frac{x}{2} - \frac{\pi}{6}\right) = 1 \Rightarrow \frac{x}{2} - \frac{\pi}{6} = \frac{\pi}{2} + 2\pi n \Rightarrow \frac{x}{2} = \frac{2\pi}{3} + 2\pi n \Rightarrow x = \frac{4\pi}{3} + 4\pi n, n \in \mathbb{Z}.$ 3. $\cos^2 x + 2\sin x + 2 = 0 \Rightarrow 1 - \sin^2 x + 2\sin x + 2 = 0 \Rightarrow \sin^2 x - 2\sin x - 3 = 0.$ Пусть $\sin x = t, |t| \le 1$: $t^2 - 2t - 3 = 0 \Rightarrow t_1 = 3$ (не подходит), $t_2 = -1.$ $\sin x = -1 \Rightarrow x = -\frac{\pi}{2} + 2\pi n, n \in \mathbb{Z}.$ 4. $6\sin^2 x - 5\sin x \cos x + \cos^2 x = 0.$ Разделим на $\cos^2 x \neq 0$: $6\text{tg}^2 x - 5\text{tg} x + 1 = 0.$ Пусть $\text{tg} x = a$: $6a^2 - 5a + 1 = 0 \Rightarrow D = 25 - 24 = 1.$ $a_1 = \frac{5+1}{12} = \frac{1}{2}; a_2 = \frac{5-1}{12} = \frac{1}{3}.$ $x = \text{arctg}\left(\frac{1}{2}\right) + \pi n; x = \text{arctg}\left(\frac{1}{3}\right) + \pi n, n \in \mathbb{Z}.$ 5. $5\sin^2 x + 2\sin x \cos x - \cos^2 x = \sin^2 x + \cos^2 x$ $4\sin^2 x + 2\sin x \cos x - 2\cos^2 x = 0 \Rightarrow 2\sin^2 x + \sin x \cos x - \cos^2 x = 0.$ Разделим на $\cos^2 x$: $2\text{tg}^2 x + \text{tg} x - 1 = 0.$ Пусть $\text{tg} x = a$: $2a^2 + a - 1 = 0 \Rightarrow a_1 = -1, a_2 = 0,5.$ $x = -\frac{\pi}{4} + \pi n; x = \text{arctg}(0,5) + \pi n, n \in \mathbb{Z}.$ 6. $\sin 3x + \cos 3x = 0 \Rightarrow \text{tg} 3x = -1 \Rightarrow 3x = -\frac{\pi}{4} + \pi n \Rightarrow x = -\frac{\pi}{12} + \frac{\pi n}{3}, n \in \mathbb{Z}.$ Отберем корни на $[0; 6]$ (учитывая $\pi \approx 3,14$): $n=1: x = \frac{3\pi}{12} = \frac{\pi}{4} \approx 0,785$ $n=2: x = \frac{7\pi}{12} \approx 1,83$ $n=3: x = \frac{11\pi}{12} \approx 2,88$ $n=4: x = \frac{15\pi}{12} = \frac{5\pi}{4} \approx 3,93$ $n=5: x = \frac{19\pi}{12} \approx 4,97$ $n=6: x = \frac{23\pi}{12} \approx 6,02$ (не входит, так как $6,02 > 6$) **Ответ:** $\frac{\pi}{4}; \frac{7\pi}{12}; \frac{11\pi}{12}; \frac{5\pi}{4}; \frac{19\pi}{12}.$