XII.2. 1) cos(x+\frac{\pi}{6})=\frac{\sqrt{3}}{2}; 2) sin(x-\frac{\pi}{6})=-\frac{1}{2}; 3) sin(\frac{\pi}{4}-x)=-\frac{\sqrt{2}}{2}; 4) cos(x-\frac{3\pi}{4})=-\frac{\sqrt{2}}{2}. XII.3. 1) sin x = \frac{1}{4}; 3) cos x = -\frac{1}{7}; XII.4. 1) sin(2-x) - \sqrt{3}=0; 3) 3 sin(x-2) = 1;

XII.2. 1) cos(x+\frac{\pi}{6})=\frac{\sqrt{3}}{2} \[x + \frac{\pi}{6} = \pm \frac{\pi}{6} + 2\pi k, k \in Z\] \[x_1 = - \frac{\pi}{6} + \frac{\pi}{6} + 2\pi k = 2\pi k, k \in Z\] \[x_2 = - \frac{\pi}{6} - \frac{\pi}{6} + 2\pi k = -\frac{\pi}{3} + 2\pi k, k \in Z\] 2) sin(x-\frac{\pi}{6})=-\frac{1}{2} \[x - \frac{\pi}{6} = -\frac{\pi}{6} + 2\pi k, k \in Z\] \[x_1 = 2\pi k, k \in Z\] \[x - \frac{\pi}{6} = \pi + \frac{\pi}{6} + 2\pi k, k \in Z\] \[x_2 = \frac{2\pi}{6} + \pi + 2\pi k = \frac{7\pi}{6} + 2\pi k, k \in Z\] 3) sin(\frac{\pi}{4}-x)=-\frac{\sqrt{2}}{2} \[\frac{\pi}{4} - x = -\frac{\pi}{4} + 2\pi k, k \in Z\] \[x_1 = \frac{\pi}{4} + \frac{\pi}{4} + 2\pi k = \frac{\pi}{2} + 2\pi k, k \in Z\] \[\frac{\pi}{4} - x = \pi + \frac{\pi}{4} + 2\pi k, k \in Z\] \[x_2 = \frac{\pi}{4} - \pi - \frac{\pi}{4} + 2\pi k = - \pi + 2\pi k, k \in Z\] 4) cos(x-\frac{3\pi}{4})=-\frac{\sqrt{2}}{2} \[x - \frac{3\pi}{4} = -\frac{\pi}{4} + 2\pi k, k \in Z\] \[x_1 = -\frac{\pi}{4} + \frac{3\pi}{4} + 2\pi k = \frac{\pi}{2} + 2\pi k, k \in Z\] \[x - \frac{3\pi}{4} = \frac{\pi}{4} + 2\pi k, k \in Z\] \[x_2 = \frac{\pi}{4} + \frac{3\pi}{4} + 2\pi k = \pi + 2\pi k, k \in Z\] XII.3. 1) sin x = \frac{1}{4}; \[x = (-1)^k arcsin(\frac{1}{4}) + \pi k, k \in Z\] 3) cos x = -\frac{1}{7}; \[x = \pm arccos(-\frac{1}{7}) + 2\pi k, k \in Z\] XII.4. 1) sin(2-x) - \sqrt{3}=0; \[sin(2-x) = \sqrt{3}\] \sqrt{3} > 1, значит решения нет. 3) 3 sin(x-2) = 1 \[sin(x-2) = \frac{1}{3}\] \[x - 2 = (-1)^k arcsin(\frac{1}{3}) + \pi k, k \in Z\] \[x = 2 + (-1)^k arcsin(\frac{1}{3}) + \pi k, k \in Z\]

Ответ: См. выше

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