Решение:
- \( 2\cos^2 3x - 1 = 0 \)
- \( \cos^2 3x = \frac{1}{2} \)
- \( \cos 3x = \pm \frac{1}{\sqrt{2}} \)
- \( 3x = \frac{\pi}{4} + \frac{\pi k}{2} \)
- \( x = \frac{\pi}{12} + \frac{\pi k}{6}, k \in \mathbb{Z} \)
- 2) \( \cos^4 5x - \sin^4 5x = 0 \)
- \( (\cos^2 5x - \sin^2 5x)(\cos^2 5x + \sin^2 5x) = 0 \)
- \( \cos 10x \cdot 1 = 0 \)
- \( 10x = \frac{\pi}{2} + \pi k \)
- \( x = \frac{\pi}{20} + \frac{\pi k}{10}, k \in \mathbb{Z} \)
- 3) \( \sqrt{3} \cos (\frac{\pi}{3} + x) + \frac{3}{2} \sin x = \frac{\sqrt{3}}{2} \)
- \( \sqrt{3} (\cos \frac{\pi}{3} \cos x - \sin \frac{\pi}{3} \sin x) + \frac{3}{2} \sin x = \frac{\sqrt{3}}{2} \)
- \( \sqrt{3} (\frac{1}{2} \cos x - \frac{\sqrt{3}}{2} \sin x) + \frac{3}{2} \sin x = \frac{\sqrt{3}}{2} \)
- \( \frac{\sqrt{3}}{2} \cos x - \frac{3}{2} \sin x + \frac{3}{2} \sin x = \frac{\sqrt{3}}{2} \)
- \( \frac{\sqrt{3}}{2} \cos x = \frac{\sqrt{3}}{2} \)
- \( \cos x = 1 \)
- \( x = 2\pi k, k \in \mathbb{Z} \)
- 4) \( 1 - 3\sin^2 3x = \sin 3x - 3(1 - \cos 3x)(1 + \cos 3x) \)
- \( 1 - 3\sin^2 3x = \sin 3x - 3(1 - \cos^2 3x) \)
- \( 1 - 3\sin^2 3x = \sin 3x - 3\sin^2 3x \)
- \( 1 = \sin 3x \)
- \( 3x = \frac{\pi}{2} + 2\pi k \)
- \( x = \frac{\pi}{6} + \frac{2\pi k}{3}, k \in \mathbb{Z} \)
Ответ: 1) \( x = \frac{\pi}{12} + \frac{\pi k}{6}, k \in \mathbb{Z} \); 2) \( x = \frac{\pi}{20} + \frac{\pi k}{10}, k \in \mathbb{Z} \); 3) \( x = 2\pi k, k \in \mathbb{Z} \); 4) \( x = \frac{\pi}{6} + \frac{2\pi k}{3}, k \in \mathbb{Z} \).