№4 Ремить уравнения a) 8in=200570-00550-08-20° 6) Su3x-sinx=0 1-cosx

а) \(sin \frac{x}{2} = 2cos70 \cdot cos50 - cos20\) Используем формулу произведения косинусов: 2cosα \(\cdot\) cosβ = cos(α + β) + cos(α - β) \(2cos70 \cdot cos50 = cos(70 + 50) + cos(70 - 50) = cos120 + cos20 = -\frac{1}{2} + cos20\) Тогда, \(sin \frac{x}{2} = -\frac{1}{2} + cos20 - cos20 = -\frac{1}{2}\) \(sin \frac{x}{2} = -\frac{1}{2}\) \(\frac{x}{2} = arcsin(-\frac{1}{2})\) \(\frac{x}{2} = -\frac{\pi}{6} + 2\pi n, n \in Z\) или \(\frac{x}{2} = \pi - (-\frac{\pi}{6}) + 2\pi k, k \in Z\) x = -\(\frac{\pi}{3}\) + 4\(\pi\)n, n \in Z или x = \(\frac{7\pi}{3}\) + 4\(\pi\)k, k \in Z

б) \(\frac{sin3x - sinx}{1 - cosx} = 0\) Используем формулу разности синусов: sinα - sinβ = 2cos((α + β)/2)sin((α - β)/2) \(\frac{2cos((3x + x)/2)sin((3x - x)/2)}{1 - cosx} = 0\) \(\frac{2cos(2x)sin(x)}{1 - cosx} = 0\) 2cos(2x)sin(x) = 0 и 1 - cosx ≠ 0 1 - cosx ≠ 0 cоsx ≠ 1 x ≠ 2\(\pi\)n, n \in Z

• cos2x = 0 2x = \(\frac{\pi}{2}\) + \(\pi\)k, k \in Z x = \(\frac{\pi}{4}\) + \(\frac{\pi}{2}\)k, k \in Z
• sinx = 0 x = \(\pi\)k, k \in Z x = \(\pi\) + 2\(\pi\)n, n \in Z и x = 2\(\pi\)n, n \in Z - посторонний корень

Ответ: а) x = -\(\frac{\pi}{3}\) + 4\(\pi\)n, n \in Z или x = \(\frac{7\pi}{3}\) + 4\(\pi\)k, k \in Z; б) x = \(\frac{\pi}{4}\) + \(\frac{\pi}{2}\)k, k \in Z, x = \(\pi\) + 2\(\pi\)n, n \in Z