4.. Решить уравнение: 1) sin 2x - sinx = 0 2) 10 cos 2x + 3cosx = 1 3) 5 sin x + cosx = 5 4) sin 4x + cos 4x = sin 2 2x - 1

Решение уравнений:

1) \[\sin^2x - \sin x = 0\] \[\sin x (\sin x - 1) = 0\] \[\sin x = 0 \quad \text{или} \quad \sin x = 1\] \[x = \pi n, n \in \mathbb{Z} \quad \text{или} \quad x = \frac{\pi}{2} + 2\pi k, k \in \mathbb{Z}\] 2) \[[10\cos^2x + 3\cos x = 1\] \[10\cos^2x + 3\cos x - 1 = 0\] Пусть \(t = \cos x\), тогда: \[10t^2 + 3t - 1 = 0\] \[D = 3^2 - 4 \cdot 10 \cdot (-1) = 9 + 40 = 49\] \[t_1 = \frac{-3 + 7}{20} = \frac{4}{20} = \frac{1}{5}\] \[t_2 = \frac{-3 - 7}{20} = \frac{-10}{20} = -\frac{1}{2}\] \[\cos x = \frac{1}{5} \quad \text{или} \quad \cos x = -\frac{1}{2}\] \[x = \pm \arccos\frac{1}{5} + 2\pi n, n \in \mathbb{Z} \quad \text{или} \quad x = \pm \frac{2\pi}{3} + 2\pi k, k \in \mathbb{Z}\] 3) \[[5\sin x + \cos x = 5\] Так как \([\sin x \le 1\) и \([\cos x \le 1\), то уравнение может выполняться только при \([\sin x = 1\) и \([\cos x = 0\). \[5 \cdot 1 + 0 = 5\] \[x = \frac{\pi}{2} + 2\pi n, n \in \mathbb{Z}\] 4) \[[\sin^4x + \cos^4x = \sin^2 2x - \frac{1}{2}\] \[((\sin^2x + \cos^2x)^2 - 2\sin^2x \cos^2x) = \sin^2 2x - \frac{1}{2}\] \[1 - 2\sin^2x \cos^2x = \sin^2 2x - \frac{1}{2}\] \[1 - \frac{1}{2} \sin^2 2x = \sin^2 2x - \frac{1}{2}\] \[\frac{3}{2} = \frac{3}{2} \sin^2 2x\] \[\sin^2 2x = 1\] \[\sin 2x = \pm 1\] \[2x = \frac{\pi}{2} + \pi n, n \in \mathbb{Z}\] \[x = \frac{\pi}{4} + \frac{\pi n}{2}, n \in \mathbb{Z}\]

Ответ: 1) \[x = \pi n, n \in \mathbb{Z}\]; \[x = \frac{\pi}{2} + 2\pi k, k \in \mathbb{Z}\] 2) \[x = \pm \arccos\frac{1}{5} + 2\pi n, n \in \mathbb{Z}\]; \[x = \pm \frac{2\pi}{3} + 2\pi k, k \in \mathbb{Z}\] 3) \[x = \frac{\pi}{2} + 2\pi n, n \in \mathbb{Z}\] 4) \[x = \frac{\pi}{4} + \frac{\pi n}{2}, n \in \mathbb{Z}\]