Question

10 класс алгебра 1 и 2 вариант

Answer 1

1 вариант

1.

$\sqrt{2} \cos(4x) - 1 = 0 \\ \cos(4x) = \frac{1}{ \sqrt{2} } \\ \cos(4x) = \frac{ \sqrt{2} }{2} \\ 4x = \pm \frac{\pi}{4} + 2\pi \: n \\ x = \pm \frac{\pi}{16} + \frac{\pi \: n}{2}$

2.

$\cos( \frac{\pi}{3} - 3x) = \frac{1}{2} \\ \\ \frac{\pi}{3} - 3x = \frac{\pi}{3} + 2\pi \: n \\ - 3x = 2\pi \: n \\ x_1= - \frac{2\pi \: n}{ 3} \\ \\ \frac{\pi}{3} - 3x = - \frac{\pi}{3} + 2\pi \: n \\ - 3x = - \frac{2\pi}{3} + 2\pi \: n \\ x_2 = \frac{2\pi}{9} - \frac{2\pi \: n}{3}$

3.

$- 2 \cos(x) = 0 \\ \cos(x) = 0 \\ x = \frac{\pi}{2} + \pi \: n$

4.

$- \sqrt{3} ctg \frac{x}{4} = 3 \\ ctg \frac{x}{4} = - \sqrt{3} \\ \frac{x}{4} = - \frac{\pi}{6} + \pi \: n \\ x = - \frac{2\pi}{3} + 4\pi \: n$

2 вариант

1.

$\cos(2x - \frac{\pi}{2} ) = \frac{ \sqrt{2} }{2} \\ \sin(2x) = \frac{ \sqrt{2} }{2} \\ \\ 2x = \frac{\pi}{4} + 2 \pi \: n \\ x_1 = \frac{\pi}{8} + \pi \: n \\ \\ 2x = \frac{3\pi}{4} + 2\pi \: n \\ x_2 = \frac{3\pi}{8} + \pi \: n$

2.

$\sin(4x + \frac{3\pi}{2} ) = 0 \\ - \cos(4x) = 0 \\ \cos(4x) = 0 \\ 4x = \frac{\pi}{2} + \pi \: n \\ x = \frac{\pi}{8} + \frac{\pi \: n}{4}$

3.

$\cos( - x) = - 1 \\ \cos(x) = - 1 \\ x = \pi + 2\pi \: n$

4.

$\sqrt{3} tg \frac{x}{2} = 1 \\ tg \frac{x}{2} = \frac{1}{ \sqrt{3} } = \frac{ \sqrt{3} }{3} \\ \frac{x}{2} = \frac{\pi}{6} + \pi \: n \\ x = \frac{\pi}{3} + 2 \pi \: n$

везде n принадлежит Z.