6) 2sin2x-4 cos x + 3 sin x - 3 = 0 7) 2cos2x - √2sin(x-π) + 1=0; 8) 4 2cos2x + cos(23/7-x)+1=0 π 9) cosx + 2cos(2x- -) = √3sin2x-1. 3 10) 5 - 8cos²x = sin 2x. 11) cos4x cosx +sin4x sinx=0. 12) 4 sinx = cos(x-5) 4 2 13) 4 sin²x + 3 cos 2x - 1 = 0. 3π 14) 4 cos²x + 8sin(+ x) + 1 = 0 15) cos 2x + √2 cos π + 1 = 0. (1+x) +1 2 16) 3cos2x - 5 sinx +1 = 0. π 17) 2 sin² x + 3√3sin( 1 + x) +4=0 2

Решение:

6) \( 2\sin 2x - 4\cos x + 3\sin x - 3 = 0 \) \( 2(2\sin x \cos x) - 4 \cos x + 3(\sin x - 1) = 0 \) \( 4\sin x \cos x - 4 \cos x + 3(\sin x - 1) = 0 \) \( 4\cos x(\sin x - 1) + 3(\sin x - 1) = 0 \) \( (4\cos x + 3)(\sin x - 1) = 0 \) \( 4\cos x + 3 = 0 \) или \( \sin x - 1 = 0 \) \( \cos x = -\frac{3}{4} \) или \( \sin x = 1 \) \( x = \pm \arccos(-\frac{3}{4}) + 2\pi n, n \in \mathbb{Z} \) или \( x = \frac{\pi}{2} + 2\pi k, k \in \mathbb{Z} \) 7) \( 2\cos 2x - \sqrt{2}\sin(x-\pi) + 1 = 0 \) \( 2\cos 2x + \sqrt{2}\sin(x) + 1 = 0 \) \( 2(1 - 2\sin^2 x) + \sqrt{2}\sin x + 1 = 0 \) \( 2 - 4\sin^2 x + \sqrt{2}\sin x + 1 = 0 \) \( -4\sin^2 x + \sqrt{2}\sin x + 3 = 0 \) \( 4\sin^2 x - \sqrt{2}\sin x - 3 = 0 \) Пусть \( t = \sin x \), тогда: \( 4t^2 - \sqrt{2}t - 3 = 0 \) \( D = (-\sqrt{2})^2 - 4 \cdot 4 \cdot (-3) = 2 + 48 = 50 \) \( t_1 = \frac{\sqrt{2} + \sqrt{50}}{2 \cdot 4} = \frac{\sqrt{2} + 5\sqrt{2}}{8} = \frac{6\sqrt{2}}{8} = \frac{3\sqrt{2}}{4} \) \( t_2 = \frac{\sqrt{2} - \sqrt{50}}{2 \cdot 4} = \frac{\sqrt{2} - 5\sqrt{2}}{8} = \frac{-4\sqrt{2}}{8} = -\frac{\sqrt{2}}{2} \) \( \sin x = \frac{3\sqrt{2}}{4} \) (не имеет решений, так как \( \frac{3\sqrt{2}}{4} > 1 \)) \( \sin x = -\frac{\sqrt{2}}{2} \) \( x = -\frac{\pi}{4} + 2\pi n, n \in \mathbb{Z} \) или \( x = \frac{5\pi}{4} + 2\pi k, k \in \mathbb{Z} \) 8) \( 2\cos 2x + 4\cos(\frac{3\pi}{2} - x) + 1 = 0 \) \( 2\cos 2x - 4\sin x + 1 = 0 \) \( 2(1 - 2\sin^2 x) - 4\sin x + 1 = 0 \) \( 2 - 4\sin^2 x - 4\sin x + 1 = 0 \) \( -4\sin^2 x - 4\sin x + 3 = 0 \) \( 4\sin^2 x + 4\sin x - 3 = 0 \) Пусть \( t = \sin x \), тогда: \( 4t^2 + 4t - 3 = 0 \) \( D = 4^2 - 4 \cdot 4 \cdot (-3) = 16 + 48 = 64 \) \( t_1 = \frac{-4 + \sqrt{64}}{2 \cdot 4} = \frac{-4 + 8}{8} = \frac{4}{8} = \frac{1}{2} \) \( t_2 = \frac{-4 - \sqrt{64}}{2 \cdot 4} = \frac{-4 - 8}{8} = \frac{-12}{8} = -\frac{3}{2} \) \( \sin x = \frac{1}{2} \) или \( \sin x = -\frac{3}{2} \) (не имеет решений, так как \( -\frac{3}{2} < -1 \)) \( x = \frac{\pi}{6} + 2\pi n, n \in \mathbb{Z} \) или \( x = \frac{5\pi}{6} + 2\pi k, k \in \mathbb{Z} \) 9) \( \cos x + 2\cos(2x - \frac{\pi}{3}) = \sqrt{3}\sin 2x - 1 \) \( \cos x + 2(\cos 2x \cos \frac{\pi}{3} + \sin 2x \sin \frac{\pi}{3}) = \sqrt{3}\sin 2x - 1 \) \( \cos x + 2(\cos 2x \cdot \frac{1}{2} + \sin 2x \cdot \frac{\sqrt{3}}{2}) = \sqrt{3}\sin 2x - 1 \) \( \cos x + \cos 2x + \sqrt{3}\sin 2x = \sqrt{3}\sin 2x - 1 \) \( \cos x + \cos 2x = -1 \) \( \cos x + 2\cos^2 x - 1 = -1 \) \( \cos x + 2\cos^2 x = 0 \) \( \cos x(1 + 2\cos x) = 0 \) \( \cos x = 0 \) или \( 1 + 2\cos x = 0 \) \( x = \frac{\pi}{2} + \pi n, n \in \mathbb{Z} \) или \( \cos x = -\frac{1}{2} \) \( x = \frac{\pi}{2} + \pi n, n \in \mathbb{Z} \) или \( x = \pm \frac{2\pi}{3} + 2\pi k, k \in \mathbb{Z} \) 10) \( 5 - 8\cos^2 x = \sin 2x \) \( 5 - 8\cos^2 x = 2\sin x \cos x \) \( 5 - 8\cos^2 x - 2\sin x \cos x = 0 \) Это уравнение сложно решить аналитически. 11) \( \cos 4x \cos x + \sin 4x \sin x = 0 \) \( \cos(4x - x) = 0 \) \( \cos(3x) = 0 \) \( 3x = \frac{\pi}{2} + \pi n, n \in \mathbb{Z} \) \( x = \frac{\pi}{6} + \frac{\pi n}{3}, n \in \mathbb{Z} \) 12) \( 4\sin^3 x = \cos(x - \frac{5\pi}{2}) \) \( 4\sin^3 x = \cos(x - \frac{\pi}{2} - 2\pi) \) \( 4\sin^3 x = \cos(x - \frac{\pi}{2}) \) \( 4\sin^3 x = \sin x \) \( 4\sin^3 x - \sin x = 0 \) \( \sin x(4\sin^2 x - 1) = 0 \) \( \sin x = 0 \) или \( 4\sin^2 x - 1 = 0 \) \( x = \pi n, n \in \mathbb{Z} \) или \( \sin^2 x = \frac{1}{4} \) \( x = \pi n, n \in \mathbb{Z} \) или \( \sin x = \pm \frac{1}{2} \) \( x = \pi n, n \in \mathbb{Z} \) или \( x = \frac{\pi}{6} + 2\pi k, k \in \mathbb{Z} \), \( x = \frac{5\pi}{6} + 2\pi m, m \in \mathbb{Z} \), \( x = -\frac{\pi}{6} + 2\pi l, l \in \mathbb{Z} \), \( x = \frac{7\pi}{6} + 2\pi p, p \in \mathbb{Z} \) 13) \( 4\sin^4 x + 3\cos 2x - 1 = 0 \) \( 4\sin^4 x + 3(1 - 2\sin^2 x) - 1 = 0 \) \( 4\sin^4 x + 3 - 6\sin^2 x - 1 = 0 \) \( 4\sin^4 x - 6\sin^2 x + 2 = 0 \) \( 2\sin^4 x - 3\sin^2 x + 1 = 0 \) Пусть \( t = \sin^2 x \), тогда: \( 2t^2 - 3t + 1 = 0 \) \( D = (-3)^2 - 4 \cdot 2 \cdot 1 = 9 - 8 = 1 \) \( t_1 = \frac{3 + \sqrt{1}}{2 \cdot 2} = \frac{3 + 1}{4} = \frac{4}{4} = 1 \) \( t_2 = \frac{3 - \sqrt{1}}{2 \cdot 2} = \frac{3 - 1}{4} = \frac{2}{4} = \frac{1}{2} \) \( \sin^2 x = 1 \) или \( \sin^2 x = \frac{1}{2} \) \( \sin x = \pm 1 \) или \( \sin x = \pm \frac{\sqrt{2}}{2} \) \( x = \frac{\pi}{2} + \pi n, n \in \mathbb{Z} \) или \( x = \frac{\pi}{4} + \pi k, k \in \mathbb{Z} \) 14) \( 4\cos^2 x + 8\sin(\frac{3\pi}{2} + x) + 1 = 0 \) \( 4\cos^2 x - 8\cos x + 1 = 0 \) Пусть \( t = \cos x \), тогда: \( 4t^2 - 8t + 1 = 0 \) \( D = (-8)^2 - 4 \cdot 4 \cdot 1 = 64 - 16 = 48 \) \( t_1 = \frac{8 + \sqrt{48}}{2 \cdot 4} = \frac{8 + 4\sqrt{3}}{8} = \frac{2 + \sqrt{3}}{2} \) \( t_2 = \frac{8 - \sqrt{48}}{2 \cdot 4} = \frac{8 - 4\sqrt{3}}{8} = \frac{2 - \sqrt{3}}{2} \) \( \cos x = \frac{2 + \sqrt{3}}{2} \) (не имеет решений, так как \( \frac{2 + \sqrt{3}}{2} > 1 \)) \( \cos x = \frac{2 - \sqrt{3}}{2} \) \( x = \pm \arccos(\frac{2 - \sqrt{3}}{2}) + 2\pi n, n \in \mathbb{Z} \) 15) \( \cos 2x + \sqrt{2}\cos(\frac{\pi}{2} + x) + 1 = 0 \) \( \cos 2x - \sqrt{2}\sin x + 1 = 0 \) \( 1 - 2\sin^2 x - \sqrt{2}\sin x + 1 = 0 \) \( -2\sin^2 x - \sqrt{2}\sin x + 2 = 0 \) \( 2\sin^2 x + \sqrt{2}\sin x - 2 = 0 \) Пусть \( t = \sin x \), тогда: \( 2t^2 + \sqrt{2}t - 2 = 0 \) \( D = (\sqrt{2})^2 - 4 \cdot 2 \cdot (-2) = 2 + 16 = 18 \) \( t_1 = \frac{-\sqrt{2} + \sqrt{18}}{2 \cdot 2} = \frac{-\sqrt{2} + 3\sqrt{2}}{4} = \frac{2\sqrt{2}}{4} = \frac{\sqrt{2}}{2} \) \( t_2 = \frac{-\sqrt{2} - \sqrt{18}}{2 \cdot 2} = \frac{-\sqrt{2} - 3\sqrt{2}}{4} = \frac{-4\sqrt{2}}{4} = -\sqrt{2} \) \( \sin x = \frac{\sqrt{2}}{2} \) или \( \sin x = -\sqrt{2} \) (не имеет решений, так как \( -\sqrt{2} < -1 \)) \( x = \frac{\pi}{4} + 2\pi n, n \in \mathbb{Z} \) или \( x = \frac{3\pi}{4} + 2\pi k, k \in \mathbb{Z} \) 16) \( 3\cos 2x - 5\sin x + 1 = 0 \) \( 3(1 - 2\sin^2 x) - 5\sin x + 1 = 0 \) \( 3 - 6\sin^2 x - 5\sin x + 1 = 0 \) \( -6\sin^2 x - 5\sin x + 4 = 0 \) \( 6\sin^2 x + 5\sin x - 4 = 0 \) Пусть \( t = \sin x \), тогда: \( 6t^2 + 5t - 4 = 0 \) \( D = 5^2 - 4 \cdot 6 \cdot (-4) = 25 + 96 = 121 \) \( t_1 = \frac{-5 + \sqrt{121}}{2 \cdot 6} = \frac{-5 + 11}{12} = \frac{6}{12} = \frac{1}{2} \) \( t_2 = \frac{-5 - \sqrt{121}}{2 \cdot 6} = \frac{-5 - 11}{12} = \frac{-16}{12} = -\frac{4}{3} \) \( \sin x = \frac{1}{2} \) или \( \sin x = -\frac{4}{3} \) (не имеет решений, так как \( -\frac{4}{3} < -1 \)) \( x = \frac{\pi}{6} + 2\pi n, n \in \mathbb{Z} \) или \( x = \frac{5\pi}{6} + 2\pi k, k \in \mathbb{Z} \) 17) \( 2\sin^2 x + 3\sqrt{3}\sin(\frac{\pi}{2} + x) + 4 = 0 \) \( 2\sin^2 x + 3\sqrt{3}\cos x + 4 = 0 \) \( 2(1 - \cos^2 x) + 3\sqrt{3}\cos x + 4 = 0 \) \( 2 - 2\cos^2 x + 3\sqrt{3}\cos x + 4 = 0 \) \( -2\cos^2 x + 3\sqrt{3}\cos x + 6 = 0 \) \( 2\cos^2 x - 3\sqrt{3}\cos x - 6 = 0 \) Пусть \( t = \cos x \), тогда: \( 2t^2 - 3\sqrt{3}t - 6 = 0 \) \( D = (-3\sqrt{3})^2 - 4 \cdot 2 \cdot (-6) = 27 + 48 = 75 \) \( t_1 = \frac{3\sqrt{3} + \sqrt{75}}{2 \cdot 2} = \frac{3\sqrt{3} + 5\sqrt{3}}{4} = \frac{8\sqrt{3}}{4} = 2\sqrt{3} \) \( t_2 = \frac{3\sqrt{3} - \sqrt{75}}{2 \cdot 2} = \frac{3\sqrt{3} - 5\sqrt{3}}{4} = \frac{-2\sqrt{3}}{4} = -\frac{\sqrt{3}}{2} \) \( \cos x = 2\sqrt{3} \) (не имеет решений, так как \( 2\sqrt{3} > 1 \)) \( \cos x = -\frac{\sqrt{3}}{2} \) \( x = \pm \frac{5\pi}{6} + 2\pi n, n \in \mathbb{Z} \)