620 1) sin² x = 1/4; 621 1) 2 cos²x - sinx + 1 = 0; 622 1) tg² x = 2; 623 1) 1+7 cos² x = 3 sin 2x; 624 1) √3 cos x + sin x = 0; 625 1) sin x - cos x = 1; 626 1) cos x = cos 3x; 627 1) cos 3x - cos 5x = sin 4x; 628 1) (tg x - √3) (2 sin x/12 +1) = 0; 2) (1-√2 cos x/4)(1+√3 tg x) = 0; 3) 2 sin (x + π/6) -1 (2 tg x + 1) = 0; 4) (1+√2 cos (x+ π/4)) (tg x - 3) = 0. 629 1) √3 sin x cos x = sin² x; 630 1) 2 sin² x = 1 + 1/3 sin 4x; 631 1) 2 sin 2x-3 (sinx + cos x) + 2 = 0;

620) 1) \[\sin^2 x = \frac{1}{4}\] \[\sin x = \pm \frac{1}{2}\] \[x = (-1)^n \frac{\pi}{6} + \pi n, \quad n \in \mathbb{Z}\] 621) 1) \[2\cos^2 x - \sin x + 1 = 0\] \[2(1 - \sin^2 x) - \sin x + 1 = 0\] \[2 - 2\sin^2 x - \sin x + 1 = 0\] \[-2\sin^2 x - \sin x + 3 = 0\] \[2\sin^2 x + \sin x - 3 = 0\] Пусть \(y = \sin x\), тогда: \[2y^2 + y - 3 = 0\] \[D = 1^2 - 4 \cdot 2 \cdot (-3) = 1 + 24 = 25\] \[y_1 = \frac{-1 + 5}{4} = 1, \quad y_2 = \frac{-1 - 5}{4} = -\frac{3}{2}\] \[\sin x = 1 \Rightarrow x = \frac{\pi}{2} + 2\pi n, \quad n \in \mathbb{Z}\] \[\sin x = -\frac{3}{2}\] (не имеет решений, т.к. \(|\sin x| \le 1\)) 622) 1) \[\tan^2 x = 2\] \[\tan x = \pm \sqrt{2}\] \[x = \arctan(\pm \sqrt{2}) + \pi n, \quad n \in \mathbb{Z}\] 623) 1) \[1 + 7\cos^2 x = 3\sin 2x\] \[1 + 7\cos^2 x = 6\sin x \cos x\] \[\sin^2 x + \cos^2 x + 7\cos^2 x = 6\sin x \cos x\] \[\sin^2 x - 6\sin x \cos x + 8\cos^2 x = 0\] Разделим обе части уравнения на \(\cos^2 x\) (если \(\cos x = 0\), то \(\sin x = 0\), что невозможно, т.к. \(\sin^2 x + \cos^2 x = 1\)): \[\tan^2 x - 6\tan x + 8 = 0\] Пусть \(y = \tan x\), тогда: \[y^2 - 6y + 8 = 0\] \[D = (-6)^2 - 4 \cdot 1 \cdot 8 = 36 - 32 = 4\] \[y_1 = \frac{6 + 2}{2} = 4, \quad y_2 = \frac{6 - 2}{2} = 2\] \[\tan x = 4 \Rightarrow x = \arctan(4) + \pi n, \quad n \in \mathbb{Z}\] \[\tan x = 2 \Rightarrow x = \arctan(2) + \pi n, \quad n \in \mathbb{Z}\] 624) 1) \[\sqrt{3} \cos x + \sin x = 0\] \[\sin x = -\sqrt{3} \cos x\] Разделим обе части уравнения на \(\cos x\) (если \(\cos x = 0\), то \(\sin x = 0\), что невозможно, т.к. \(\sin^2 x + \cos^2 x = 1\)): \[\tan x = -\sqrt{3}\] \[x = -\frac{\pi}{3} + \pi n, \quad n \in \mathbb{Z}\] 625) 1) \[\sin x - \cos x = 1\] \[\sin x = \cos x + 1\] \[\sin^2 x = (\cos x + 1)^2\] \[1 - \cos^2 x = \cos^2 x + 2\cos x + 1\] \[2\cos^2 x + 2\cos x = 0\] \[2\cos x(\cos x + 1) = 0\] \[\cos x = 0 \Rightarrow x = \frac{\pi}{2} + \pi n, \quad n \in \mathbb{Z}\] \[\cos x = -1 \Rightarrow x = \pi + 2\pi n, \quad n \in \mathbb{Z}\] Проверим: \[x = \frac{\pi}{2} + \pi n\] Если \(n = 2k\), то \(x = \frac{\pi}{2} + 2\pi k\), тогда \(\sin x = 1\) и \(\cos x = 0\), следовательно \(1 - 0 = 1\) (верно). Если \(n = 2k+1\), то \(x = \frac{\pi}{2} + \pi(2k+1) = \frac{3\pi}{2} + 2\pi k\), тогда \(\sin x = -1\) и \(\cos x = 0\), следовательно \(-1 - 0 = 1\) (неверно). \[x = \pi + 2\pi n\] Тогда \(\sin x = 0\) и \(\cos x = -1\), следовательно \(0 - (-1) = 1\) (верно). 626) 1) \[\cos x = \cos 3x\] \[\cos 3x - \cos x = 0\] \[-2 \sin \frac{3x + x}{2} \sin \frac{3x - x}{2} = 0\] \[-2 \sin 2x \sin x = 0\] \[\sin 2x = 0 \Rightarrow 2x = \pi n \Rightarrow x = \frac{\pi n}{2}, \quad n \in \mathbb{Z}\] \[\sin x = 0 \Rightarrow x = \pi n, \quad n \in \mathbb{Z}\] 627) 1) \[\cos 3x - \cos 5x = \sin 4x\] \[-2 \sin \frac{3x + 5x}{2} \sin \frac{3x - 5x}{2} = \sin 4x\] \[-2 \sin 4x \sin (-x) = \sin 4x\] \[2 \sin 4x \sin x = \sin 4x\] \[2 \sin 4x \sin x - \sin 4x = 0\] \[\sin 4x(2 \sin x - 1) = 0\] \[\sin 4x = 0 \Rightarrow 4x = \pi n \Rightarrow x = \frac{\pi n}{4}, \quad n \in \mathbb{Z}\] \[2 \sin x - 1 = 0 \Rightarrow \sin x = \frac{1}{2} \Rightarrow x = (-1)^n \frac{\pi}{6} + \pi n, \quad n \in \mathbb{Z}\] 628) 1) \[(\tan x - \sqrt{3})\left(2 \sin \frac{x}{12} + 1\right) = 0\] \[\tan x - \sqrt{3} = 0 \Rightarrow \tan x = \sqrt{3} \Rightarrow x = \frac{\pi}{3} + \pi n, \quad n \in \mathbb{Z}\] \[2 \sin \frac{x}{12} + 1 = 0 \Rightarrow \sin \frac{x}{12} = -\frac{1}{2} \Rightarrow \frac{x}{12} = (-1)^n \left(-\frac{\pi}{6}\right) + \pi n \Rightarrow x = (-1)^{n+1} 2\pi + 12\pi n, \quad n \in \mathbb{Z}\] 629) 1) \[\sqrt{3} \sin x \cos x = \sin^2 x\] \[\sin x(\sqrt{3} \cos x - \sin x) = 0\] \[\sin x = 0 \Rightarrow x = \pi n, \quad n \in \mathbb{Z}\] \[\sqrt{3} \cos x - \sin x = 0 \Rightarrow \sin x = \sqrt{3} \cos x \Rightarrow \tan x = \sqrt{3} \Rightarrow x = \frac{\pi}{3} + \pi n, \quad n \in \mathbb{Z}\] 630) 1) \[2\sin^2 x = 1 + \frac{1}{3} \sin 4x\] \[2\sin^2 x = 1 + \frac{2}{3} \sin 2x \cos 2x\] \[1 - \cos 2x = 1 + \frac{2}{3} \sin 2x \cos 2x\] \[\cos 2x + \frac{2}{3} \sin 2x \cos 2x = 0\] \[\cos 2x\left(1 + \frac{2}{3} \sin 2x\right) = 0\] \[\cos 2x = 0 \Rightarrow 2x = \frac{\pi}{2} + \pi n \Rightarrow x = \frac{\pi}{4} + \frac{\pi n}{2}, \quad n \in \mathbb{Z}\] \[1 + \frac{2}{3} \sin 2x = 0 \Rightarrow \sin 2x = -\frac{3}{2}\] (не имеет решений, т.к. \(|\sin 2x| \le 1\)) 631) 1) \[2\sin 2x - 3(\sin x + \cos x) + 2 = 0\] \[4\sin x \cos x - 3(\sin x + \cos x) + 2 = 0\] Пусть \(y = \sin x + \cos x\), тогда \(y^2 = \sin^2 x + 2\sin x \cos x + \cos^2 x = 1 + 2\sin x \cos x\), следовательно \(2\sin x \cos x = y^2 - 1\). Тогда: \[2(y^2 - 1) - 3y + 2 = 0\] \[2y^2 - 2 - 3y + 2 = 0\] \[2y^2 - 3y = 0\] \[y(2y - 3) = 0\] \[y = 0 \Rightarrow \sin x + \cos x = 0 \Rightarrow \tan x = -1 \Rightarrow x = -\frac{\pi}{4} + \pi n, \quad n \in \mathbb{Z}\] \[2y - 3 = 0 \Rightarrow y = \frac{3}{2} \Rightarrow \sin x + \cos x = \frac{3}{2}\] \[\sqrt{2} \sin\left(x + \frac{\pi}{4}\right) = \frac{3}{2}\] \[\sin\left(x + \frac{\pi}{4}\right) = \frac{3}{2\sqrt{2}} = \frac{3\sqrt{2}}{4} \approx 1.06\] (не имеет решений, т.к. \(|\sin x| \le 1\))