Tìm x

To solve the equations in your image, let's go through them one by one.

### Bài 1:
1. **\( 2\cos^2 x + \sqrt{2} = 0 \)**
- Rearranging gives \( \cos^2 x = -\frac{\sqrt{2}}{2} \) (no solutions).

2. **\( 2\sin x - \sqrt{2} = 0 \)**
- This simplifies to \( \sin x = \frac{\sqrt{2}}{2} \).
- Solutions: \( x = \frac{\pi}{4} + k\pi \) (where \( k \) is an integer).

3. **\( \cos 5x - 1 = 0 \)**
- Thus, \( \cos 5x = 1 \) implies \( 5x = 2k\pi \).
- Solutions: \( x = \frac{2k\pi}{5} \).

4. **\( \sin 5x + 1 = 0 \)**
- So, \( \sin 5x = -1 \) gives \( 5x = \frac{3\pi}{2} + 2k\pi \).
- Solutions: \( x = \frac{3\pi}{10} + \frac{2k\pi}{5} \).

### Bài 2:
1. **Given \( \sin \alpha = \frac{5}{\sqrt{34}} \)**, then \( \cos \alpha = \sqrt{1 - \left(\frac{5}{\sqrt{34}}\right)^2} = \frac{3}{\sqrt{34}} \).

2. **If \( \sin \alpha = \frac{2}{3}, \cos \alpha = \sqrt{1 - \left(\frac{2}{3}\right)^2} = \frac{\sqrt{5}}{3} \)**.

#### a) Find \( \sin 2\alpha, \cos 2\alpha \):
- \( \sin 2\alpha = 2\sin \alpha \cos \alpha = 2 \cdot \frac{2}{3} \cdot \frac{\sqrt{5}}{3} = \frac{4\sqrt{5}}{9} \).
- \( \cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha = \frac{5}{9} - \frac{4}{9} = \frac{1}{9} \).

#### b) Find \( \sin\left(\alpha - \frac{\pi}{2}\right) \):
- \( \sin\left(\alpha - \frac{\pi}{2}\right) = -\cos \alpha = -\frac{\sqrt{5}}{3} \).

#### c) Find \( \cos\left(\alpha + \frac{\pi}{2}\right) \):
- \( \cos\left(\alpha + \frac{\pi}{2}\right) = -\sin \alpha = -\frac{2}{3} \).

Feel free to ask if you need specific clarifications on these steps!