Question

С решением, пожалуйста:

Answer 1

Ответ:

$2 \sin {}^{2} (x) - \sin(2x) + \sin(x) = \cos(x) \\ 2 \sin {}^{2} (x) - 2 \sin(x) \cos(x) + \sin(x) - \cos(x) = 0 \\ 2 \sin(x) ( \sin(x) - \cos(x)) + ( \sin(x) - \cos(x)) = 0 \\ ( \sin(x) - \cos(x)) (2 \sin(x) + 1) = 0 \\ \\ \sin(x) - \cos(x) = 0 \\ | \div \cos(x) \ne0 \\ tgx - 1 = 0 \\ tgx = 1 \\ x_1 = \frac{\pi}{4} + \pi \: n \\ \\ 2 \sin(x) + 1 = 0 \\ \sin(x) = - \frac{1}{2} \\ x_2 = - \frac{\pi}{6} + 2 \pi \: n \\ x_3 = - \frac{5\pi}{6} + 2\pi \: n \\ n\\ \in \:$

На промежутке:

$x_1 = - 2\pi - \frac{\pi}{6} = - \frac{13\pi}{6} \\ x_2 = - 3\pi + \ \frac{\pi}{6} = - \frac{17\pi}{6} \\ x_3 = - \frac{11\pi}{4} \\ x_4 = - \frac{15\pi}{4}$

Сумма корней:

$- \frac{13\pi}{6} - \frac{17\pi}{6} - \frac{11\pi}{4} - \frac{15\pi}{4} = \\ = \frac{ - 23 - 34 - 33 - 45}{12} \pi = - \frac{135\pi}{12} = - 11.25\pi$

Ответ: -11,25