Question

4.) 2cos²x+√3cosx=0, sin x меньше 0.

Answer 1

$\left \{ {{2cos^2x+\sqrt3cosx=0} \atop {sinx\ \textless \ 0\qquad \quad }} \right. \\\\2cos^2x+\sqrt3cosx=0\\\\cosx(2cosx+\sqrt3)=0\\\\1)\; \; cosx=0\; ,\; \; x= \frac{\pi }{2}+\pi n,\; n\in Z\\\\2)\; \; cosx=-\frac{\sqrt3}{2}\; ,\; \; x=\pm (\pi -\frac{\pi}{6})+2\pi m=\pm \frac{5\pi }{6}+2\pi m,\; m\in Z\\\\3)\; \; sinx\ \textless \ 0\; \; ,\; \; -\pi +2\pi k\ \textless \ x\ \textless \ 2\pi k\; ,\; k\in Z\; \; \Rightarrow \; \; \\\\x_1=-\frac{\pi}{2}+2\pi n\; ,\; n\in Z\\\\x_2=-\frac{5\pi}{6}+2\pi m,\; m\in Z$