Question

помогите срочно математика ​

Answer 1

Ответ:

1

$\sqrt{3} \sin(x) + \cos(x) 0 \\ \sqrt{3} \sin(x) = \cos(x) \\ | \div \cos(x) \ne0 \\ \sqrt{3} tgx = 1 \\ tgx = \frac{1}{ \sqrt{3} } = \frac{ \sqrt{3} }{3} \\ x = \frac{\pi}{6} + \pi \: n$

n принадлежит Z.

2

$\sin(2x) + 2 \cos(2x) \\ \sin(2x) = - 2\cos(2x) \\ | \div \cos(2x) \ne0 \\ tg(2x) = - 2 \\ 2x = - arctg(2) + \pi \: n \\ x = - \frac{1}{2} arctg(2)+ \frac{\pi \: n}{2}$

n принадлежит Z.

3

$6 \sin {}^{2} (x) + 4 \sin(x) \cos(x) = 1 \\ 6 \sin {}^{2} (x) + 4 \sin(x ) \cos(x) = \sin {}^{2} (x) + \cos {}^{2} (x) \\ 5 \sin {}^{2} (x) + 4 \sin(x) \cos(x) - \cos {}^{2} (x) = 0 \\ | \div \cos {}^{2} (x) \ne0 \\ 5 {tg}^{2} x + 4tgx - 1 = 0 \\ \\ tgx = t \\ \\5 t {}^{2} + 4 t - 1 = 0\\ D =16 + 20 = 36 \\ t_1 = \frac{ - 4 + 6}{10} = \frac{1}{5} = 0.2 \\ t_2 = - 1 \\ \\ tgx = 0.2 \\ \\ x_1 = arctg(0.2) + \pi \: n \\ \\ tgx = - 1 \\ x_2 = - \frac{\pi}{4} + \pi \: n$

n принадлежит Z