5 sin x=12. √2 2cosx = √3 √2cos(2x-5-1 sin x = -1 2 tg (-x) = 1 6 3x sin(++1=0 sin x = 0 1 cos x = = 2 2cosx = √2 tg (-2x)=- 3 7 1 √5

Ответ: Решение тригонометрических уравнений.

1. sin x = -1

   \[x = -\frac{\pi}{2} + 2\pi k, k \in \mathbb{Z}\]

2. \[sin x = \frac{\sqrt{2}}{2}\]

   \[x = \frac{\pi}{4} + 2\pi k, k \in \mathbb{Z}\]

   \[x = \frac{3\pi}{4} + 2\pi k, k \in \mathbb{Z}\]

3. \[2 \cos x = -\sqrt{3}\]

   \[\cos x = -\frac{\sqrt{3}}{2}\]

   \[x = \frac{5\pi}{6} + 2\pi k, k \in \mathbb{Z}\]

   \[x = -\frac{5\pi}{6} + 2\pi k, k \in \mathbb{Z}\]

4. \[\sqrt{2} \cos \left(2 x-\frac{\pi}{3}\right)=1\]

   \[\cos \left(2 x-\frac{\pi}{3}\right)=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\]

   \[2 x-\frac{\pi}{3}=\frac{\pi}{4}+2 \pi k, k \in \mathbb{Z}\]

   \[2 x=\frac{\pi}{4}+\frac{\pi}{3}+2 \pi k\]

   \[2 x=\frac{7 \pi}{12}+2 \pi k\]

   \[x=\frac{7 \pi}{24}+\pi k, k \in \mathbb{Z}\]

   \[2 x-\frac{\pi}{3}=-\frac{\pi}{4}+2 \pi k\]

   \[2 x=-\frac{\pi}{4}+\frac{\pi}{3}+2 \pi k\]

   \[2 x=\frac{\pi}{12}+2 \pi k\]

   \[x=\frac{\pi}{24}+\pi k, k \in \mathbb{Z}\]

5. \[\operatorname{tg}(\pi-x)=1\]

   \[\pi-x=\frac{\pi}{4}+\pi k, k \in \mathbb{Z}\]

   \[-x=\frac{\pi}{4}-\pi+\pi k\]

   \[-x=-\frac{3 \pi}{4}+\pi k\]

   \[x=\frac{3 \pi}{4}-\pi k, k \in \mathbb{Z}\]

   \[x=\frac{3 \pi}{4}+\pi k, k \in \mathbb{Z}\]

1. sin x = 0

   \[x = \pi k, k \in \mathbb{Z}\]

2. \[\cos x = \frac{1}{2}\]

   \[x = \frac{\pi}{3} + 2\pi k, k \in \mathbb{Z}\]

   \[x = -\frac{\pi}{3} + 2\pi k, k \in \mathbb{Z}\]

3. \[2 \cos x = \sqrt{2}\]

   \[\cos x = \frac{\sqrt{2}}{2}\]

   \[x = \frac{\pi}{4} + 2\pi k, k \in \mathbb{Z}\]

   \[x = -\frac{\pi}{4} + 2\pi k, k \in \mathbb{Z}\]

4. \[\operatorname{tg}(-2 x)=-\frac{\sqrt{3}}{3}\]

   \[-2 x=-\frac{\pi}{6}+\pi k, k \in \mathbb{Z}\]

   \[x=\frac{\pi}{12}-\frac{\pi k}{2}, k \in \mathbb{Z}\]

5. \[\sin \left(\frac{3 x}{2}+\frac{\pi}{3}\right)+1=0\]

   \[\sin \left(\frac{3 x}{2}+\frac{\pi}{3}\right)=-1\]

   \[\frac{3 x}{2}+\frac{\pi}{3}=-\frac{\pi}{2}+2 \pi k, k \in \mathbb{Z}\]

   \[\frac{3 x}{2}=-\frac{\pi}{2}-\frac{\pi}{3}+2 \pi k\]

   \[\frac{3 x}{2}=-\frac{5 \pi}{6}+2 \pi k\]

   \[x=-\frac{5 \pi}{9}+\frac{4 \pi k}{3}, k \in \mathbb{Z}\]

Ответ: Решение тригонометрических уравнений.