Question

cos(2x+П/6)+(корень из 3)/2=0

Answer 1

$cos(2x+frac{pi}{6})=-frac{sqrt{3}}{2}$

$2x+frac{pi}{6}=_-^+(pi-frac{pi}{6})+2pi n$

$x=_-^+frac{5pi}{12} -frac{pi}{12}+pi n$

x$_1$=$-frac{5pi}{12} -frac{pi}{12}+pi n=-frac{pi}{2}+pi n$

$x_2=frac{5pi}{12} -frac{pi}{12}+pi n=frac{pi}{3}+pi n$