Question

Срочно помогите, нужно решить уравнения

Answer 1

Ответ:

1.

$tgx - 2 = 0 \\ tgx = 2 \\ x = arctg(2) + \pi \: n$

2.

$\cos(x) = - \frac{ \sqrt{3} }{2} \\ x = \pm \frac{5\pi}{6} + 2 \pi \: n$

3.

$4 \sin {}^{2} (x) - 4\sin(x) - 3 = 0 \\ \\ \sin(x) = t \\ \\ 4t {}^{2} - 4 t - 3 = 0 \\ D = 16 + 48 = 64 \\ t_1 = \frac{4 + 8}{8} = \frac{12}{8} = 1.5 \\ t_2 = - \frac{1}{2} \\ \\ \sin(x) = 1.5$

нет корней

$\sin(x) = - \frac{1}{2} \\ x_1 = - \frac{\pi}{6} + 2\pi \: n \\ x_2 = - \frac{5\pi}{6} + 2\pi \: n$

4.

${tg}^{2} x - 4tgx + 3 = 0\\ \\ tgx = t \\ \\ t {}^{2} - 4 t + 3 = 0 \\ D= 16 - 12 = 4 \\ t_1 = \frac{4 + 2}{2} = 3 \\ t_2 = 1 \\ \\ tgx = 3 \\ x_1 = arctg(3) + \pi \: n \\ \\ tgx = 1 \\ x_2 = \frac{\pi}{4} + \pi \: n$

5.

$3 \cos(x) - 2 \sin {}^{2} (x) = 0 \\ 3 \cos(x) - 2(1 - \cos {}^{2} (x)) = 0 \\ 3 \cos(x) - 2 + 2 \cos {}^{2} (x) = 0 \\ 2 \cos {}^{2} (x) + 3 \cos(x) - 2 = 0 \\ \\ \cos(x) = t \\ \\ 2t {}^{2} + 3t - 2 = 0 \\ D = 9 + 16 = 25 \\ t_1 = \frac{ - 3 + 5}{4} = \frac{1}{2} \\ t_2 = - 2 \\ \\ \cos(x) = \frac{1}{2} \\ x = \pm \frac{\pi}{3} + 2 \pi \: n \\ \\ \cos(x) = - 2$

нет корней

6.

$3 \sin {}^{2} (x) - 8 \sin(x) \cos {}^{2} (x) + 5 \cos(x) = 0 \\ | \div \cos {}^{2} (x) \\ 3 {tg}^{2} x - 8tgx + 5 = 0\\ \\ tgx = t \\ \\ 3t {}^{2} - 8t + 5 = 0 \\ D = 64 - 60 = 4 \\ t_1 = \frac{8 + 2}{6} = \frac{5}{3} \\ t_2 = 1 \\ \\ tgx = \frac{5}{3} \\ x_1 = arctg( \frac{5}{3} ) + \pi \: n \\ \\ tgx = 1 \\ x_2 = \frac{\pi}{4} + \pi \: n$

везде n принадлежит Z.