Решить уравнение (656-665). 656 2) cos (6 + 3x) = -√2/2; 4) 2 cos (π/3 - 3x) - √3 = 0; 657 2) 1 - sin (x/2 + π/3) = 0; 4) 5 sin (2x - 1) - 2 = 0; 659 2) tg (3x - π/4) = 1/√3; 4) 1 - tg (x + π/7) = 0; 660 2) 3 sin²x - 5 sin x - 2 = 0; 4) 6 cos² x + 7 cos x - 3 = 0. 661 2) 8 cos² x - 12 sin x + 7 = 0.

Решение:

656.

1. 2)

   \( \cos (6 + 3x) = -\frac{\sqrt{2}}{2} \)
   \( 6 + 3x = \pm \frac{3\pi}{4} + 2\pi k, k \in \mathbb{Z} \)
   \( 3x = -6 \pm \frac{3\pi}{4} + 2\pi k \)
   \( x = -2 \pm \frac{\pi}{4} + \frac{2\pi k}{3} \)

2. 4)

   \( 2 \cos \left(\frac{\pi}{3} - 3x\right) - \sqrt{3} = 0 \)
   \( \cos \left(\frac{\pi}{3} - 3x\right) = \frac{\sqrt{3}}{2} \)
   \( \frac{\pi}{3} - 3x = \pm \frac{\pi}{6} + 2\pi k, k \in \mathbb{Z} \)
   \( -3x = -\frac{\pi}{3} \pm \frac{\pi}{6} + 2\pi k \)
   \( 3x = \frac{\pi}{3} \mp \frac{\pi}{6} - 2\pi k \)
   \( 3x_1 = \frac{\pi}{3} - \frac{\pi}{6} - 2\pi k = \frac{\pi}{6} - 2\pi k \Rightarrow x_1 = \frac{\pi}{18} - \frac{2\pi k}{3} \)
   \( 3x_2 = \frac{\pi}{3} + \frac{\pi}{6} - 2\pi k = \frac{3\pi}{6} - 2\pi k = \frac{\pi}{2} - 2\pi k \Rightarrow x_2 = \frac{\pi}{6} - \frac{2\pi k}{3} \)

657.

1. 2)

   \( 1 - \sin \left(\frac{x}{2} + \frac{\pi}{3}\right) = 0 \)
   \( \sin \left(\frac{x}{2} + \frac{\pi}{3}\right) = 1 \)
   \( \frac{x}{2} + \frac{\pi}{3} = \frac{\pi}{2} + 2\pi k \)
   \( \frac{x}{2} = \frac{\pi}{2} - \frac{\pi}{3} + 2\pi k = \frac{3\pi - 2\pi}{6} + 2\pi k = \frac{\pi}{6} + 2\pi k \)
   \( x = \frac{\pi}{3} + 4\pi k \)

2. 4)

   \( 5 \sin (2x - 1) - 2 = 0 \)
   \( \sin (2x - 1) = \frac{2}{5} \)
   \( 2x - 1 = \arcsin \left(\frac{2}{5}\right) + 2\pi k \quad \text{или} \quad 2x - 1 = \pi - \arcsin \left(\frac{2}{5}\right) + 2\pi k \)
   \( 2x = 1 + \arcsin \left(\frac{2}{5}\right) + 2\pi k \Rightarrow x = \frac{1}{2} + \frac{1}{2} \arcsin \left(\frac{2}{5}\right) + \pi k \)
   \( 2x = 1 + \pi - \arcsin \left(\frac{2}{5}\right) + 2\pi k \Rightarrow x = \frac{1}{2} + \frac{\pi}{2} - \frac{1}{2} \arcsin \left(\frac{2}{5}\right) + \pi k \)

659.

1. 2)

   \( \tan \left(3x - \frac{\pi}{4}\right) = \frac{1}{\sqrt{3}} \)
   \( 3x - \frac{\pi}{4} = \frac{\pi}{6} + \pi k \)
   \( 3x = \frac{\pi}{4} + \frac{\pi}{6} + \pi k = \frac{3\pi + 2\pi}{12} + \pi k = \frac{5\pi}{12} + \pi k \)
   \( x = \frac{5\pi}{36} + \frac{\pi k}{3} \)

2. 4)

   \( 1 - \tan \left(x + \frac{\pi}{7}\right) = 0 \)
   \( \tan \left(x + \frac{\pi}{7}\right) = 1 \)
   \( x + \frac{\pi}{7} = \frac{\pi}{4} + \pi k \)
   \( x = \frac{\pi}{4} - \frac{\pi}{7} + \pi k = \frac{7\pi - 4\pi}{28} + \pi k = \frac{3\pi}{28} + \pi k \)

660.

1. 2)

   \( 3 \sin^2 x - 5 \sin x - 2 = 0 \)
   Пусть \( y = \sin x \). \( 3y^2 - 5y - 2 = 0 \)
   \( y = \frac{5 \pm \sqrt{(-5)^2 - 4(3)(-2)}}{2(3)} = \frac{5 \pm \sqrt{25 + 24}}{6} = \frac{5 \pm \sqrt{49}}{6} = \frac{5 \pm 7}{6} \)
   \( y_1 = \frac{5 + 7}{6} = \frac{12}{6} = 2 \) — решений нет, так как \( \sin x \le 1 \).
   \( y_2 = \frac{5 - 7}{6} = \frac{-2}{6} = -\frac{1}{3} \)
   \( \sin x = -\frac{1}{3} \Rightarrow x = \arcsin \left(-\frac{1}{3}\right) + 2\pi k \quad \text{или} \quad x = \pi - \arcsin \left(-\frac{1}{3}\right) + 2\pi k \)

2. 4)

   \( 6 \cos^2 x + 7 \cos x - 3 = 0 \)
   Пусть \( y = \cos x \). \( 6y^2 + 7y - 3 = 0 \)
   \( y = \frac{-7 \pm \sqrt{7^2 - 4(6)(-3)}}{2(6)} = \frac{-7 \pm \sqrt{49 + 72}}{12} = \frac{-7 \pm \sqrt{121}}{12} = \frac{-7 \pm 11}{12} \)
   \( y_1 = \frac{-7 + 11}{12} = \frac{4}{12} = \frac{1}{3} \)
   \( \cos x = \frac{1}{3} \Rightarrow x = \pm \arccos \left(\frac{1}{3}\right) + 2\pi k \)
   \( y_2 = \frac{-7 - 11}{12} = \frac{-18}{12} = -\frac{3}{2} \) — решений нет, так как \( \cos x \ge -1 \).

661.

1. 2)

   \( 8 \cos^2 x - 12 \sin x + 7 = 0 \)
   \( 8 (1 - \sin^2 x) - 12 \sin x + 7 = 0 \)
   \( 8 - 8 \sin^2 x - 12 \sin x + 7 = 0 \)
   \( -8 \sin^2 x - 12 \sin x + 15 = 0 \)
   \( 8 \sin^2 x + 12 \sin x - 15 = 0 \)
   Пусть \( y = \sin x \). \( 8y^2 + 12y - 15 = 0 \)
   \( y = \frac{-12 \pm \sqrt{12^2 - 4(8)(-15)}}{2(8)} = \frac{-12 \pm \sqrt{144 + 480}}{16} = \frac{-12 \pm \sqrt{624}}{16} = \frac{-12 \pm \sqrt{16 \cdot 39}}{16} = \frac{-12 \pm 4\sqrt{39}}{16} = \frac{-3 \pm \sqrt{39}}{4} \)
   \( y_1 = \frac{-3 + \sqrt{39}}{4} \approx \frac{-3 + 6.24}{4} \approx \frac{3.24}{4} = 0.81 \) (Решение есть, так как \( -1 \le y_1 \le 1 \))
   \( \sin x = \frac{-3 + \sqrt{39}}{4} \Rightarrow x = \arcsin \left(\frac{-3 + \sqrt{39}}{4}\right) + 2\pi k \quad \text{или} \quad x = \pi - \arcsin \left(\frac{-3 + \sqrt{39}}{4}\right) + 2\pi k \)
   \( y_2 = \frac{-3 - \sqrt{39}}{4} \approx \frac{-3 - 6.24}{4} \approx \frac{-9.24}{4} = -2.31 \) (Решений нет, так как \( y_2 < -1 \).

Ответ: 656. 2) \( x = -2 \pm \frac{\pi}{4} + \frac{2\pi k}{3} \); 4) \( x = \frac{\pi}{18} - \frac{2\pi k}{3} \), \( x = \frac{\pi}{6} - \frac{2\pi k}{3} \). 657. 2) \( x = \frac{\pi}{3} + 4\pi k \); 4) \( x = \frac{1}{2} + \frac{1}{2} \arcsin \left(\frac{2}{5}\right) + \pi k \), \( x = \frac{1}{2} + \frac{\pi}{2} - \frac{1}{2} \arcsin \left(\frac{2}{5}\right) + \pi k \). 659. 2) \( x = \frac{5\pi}{36} + \frac{\pi k}{3} \); 4) \( x = \frac{3\pi}{28} + \pi k \). 660. 2) \( x = \arcsin \left(-\frac{1}{3}\right) + 2\pi k \), \( x = \pi - \arcsin \left(-\frac{1}{3}\right) + 2\pi k \); 4) \( x = \pm \arccos \left(\frac{1}{3}\right) + 2\pi k \). 661. 2) \( x = \arcsin \left(\frac{-3 + \sqrt{39}}{4}\right) + 2\pi k \), \( x = \pi - \arcsin \left(\frac{-3 + \sqrt{39}}{4}\right) + 2\pi k \).