Question

Помогите. Задание прикреплено. Решение А тоже прикрепляю. Нужно Б.

Answer 1

$8sin^2x+2\sqrt3\, cosx+1=0\\\\8(1-cos^2x)+2\sqrt3\, cosx+1=0\\\\8cos^2x-2\sqrt3\, cosx-9=0\\\\D/4=3+72=75\ \ ,\ \ cosx=\dfrac{\sqrt3\pm 5\sqrt3}{8}\\\\a)\ \ cosx=-\dfrac{\sqrt3}{2}\ \ ,\ \ x=\pm \dfrac{5\pi }{6}+2\pi n\ ,\ n\in Z\\\\b)\ \ cosx=\dfrac{3\sqrt3}{4}\approx 1,299>1\ \ \Rightarrow \ \ x\in \varnothing \\\\\\c)\ \ x\in \Big[\, -\dfrac{7\pi}{2}\ ;\ -2\pi \ \Big]$

$-\dfrac{7\pi}{2}\leq -\dfrac{5\pi}{6}+2\pi n\leq -2\pi \ \qquad \ \ \ -\dfrac{7\pi}{2}\leq \dfrac{5\pi }{6}+2\pi n\leq -2\pi \\\\-\dfrac{7}{2}\leq -\dfrac{5}{6}+2n\leq -2\ \ \qquad \qquad \ \ -\dfrac{7}{2}\leq \dfrac{5}{6}+2n\leq -2\\\\-\dfrac{8}{3}\leq 2n\leq -\dfrac{7}{6}\ \ \ \qquad \qquad \qquad \ \ -\dfrac{13}{3}\leq 2n\leq -\dfrac{17}{6}\\\\-1\dfrac{1}{3}\leq n\leq -\dfrac{7}{12}\ \ \qquad \qquad \qquad \ \ \ \ -2\dfrac{1}{6}\leq n\leq -1\dfrac{5}{12}$

${}\ \ \ n=-1\in Z\qquad \qquad \qquad \qquad \ \qquad \qquad \ n=-2\in Z\\\\x=-\dfrac{5\pi}{6}-2\pi =-\dfrac{17\pi}{6}\qquad \qquad \qquad \ \ x=\dfrac{5\pi}{6}-4\pi =-\dfrac{19\pi}{6}}\\\\Otvet:\ \ 1)\ x=\pm \dfrac{5\pi}{6}+2\pi n\ ,\ n\in Z\ \ ;\ \ 2)\ x=-\dfrac{17\pi}{6}\ ,\ -\dfrac{19\pi}{6}\in \Big[\, -\dfrac{7\pi}{2}\, ;\, -2\pi \, \Big]\ .$