Question

Решите пожалуйста 2 вариант,первые 4 задания.

Answer 1

Ответ:

1.

$2 \sin {}^{2} (x) - \sin(x) - 1 = 0 \\ \\ \sin(x ) = t \\2 t {}^{2} - t - 1 = 0 \\ D = 1 + 8 = 9 \\ t_1 = \frac{1 + 3}{4} = 1 \\ t_2 = - \frac{1}{2} \\ \\ \sin(x) = 1 \\ \\ x_1 = \frac{\pi}{2} + 2 \pi \: n \\ \\ \sin(x) = - \frac{1}{2} \\ x_2 = - \frac{\pi}{6} + 2\pi \: n \\ x_3= - \frac{5\pi}{6} + 2\pi \: n$

n принадлежит Z.

2.

$2 \cos {}^{2} (x) - \cos(x) - 1 = 0 \\ \\ \cos(x) = t \\ \\ 2t {}^{2} - t - 1 = 0 \\ t_1 = 1 \\ t€2 = - \frac{1}{2} \\ \\ \cos(x) = 1 \\ x_1 = 2\pi \: n \\ \\ \cos(x) = - \frac{1}{2} \\ x_2 = \pm \frac{2\pi}{3} + 2 \pi \: n$

n принадлежит Z.

3.

$\sqrt{3} \cos(2x) + \sin(2x) = 0 \: \: \: | \times \frac{1}{2} \\ \frac{ \sqrt{3} }{2} \cos(2x) + \frac{1}{2} \sin(2x) = 0 \\ \sin( \frac{\pi}{3} ) \cos(2x) + \cos( \frac{\pi}{3} ) \sin(2x) = 0 \\ \sin( \frac{\pi}{3} + 2x ) = 0 \\ \frac{\pi}{3} + 2x = \pi \: n \\ 2x = - \frac{\pi}{3} + \pi \: n \\ x = - \frac{\pi}{6} + \frac{\pi \: n}{2}$

n принадлежит Z.

4.

$\sin {}^{2} (x) + \sin( x) \cos(x) = 2\cos {}^{2} (x) \\ | \div \cos {}^{2} (x) \ne0 \\ {tg}^{2} x + tgx = 2 \\ {tg}^{2} x + tgx - 2 = 0 \\ \\ tgx = t \\ \\ t {}^{2} + t - 2 = 0 \\ D = 1 + 8 = 9 \\ t_1 = \frac{ - 1 + 3}{2} = 1 \\ t_2 = - 2 \\ \\ tgx = 1 \\ x_1 = \frac{\pi}{4} + \pi \: n\\ tgx = - 2 \\ x_2 = - \arctg(2) + \pi \: n$

n принадлежит Z