Question 790215
The graph of {{{cos(x)}}} looks like this:
{{{graph(750,250,-0.5,7,-1.25,1.25,cos(x))}}}
Sine and cosine range from -1 to 1, and have a period of {{{2pi}}} (4 quadrants).
Over a period, they go full circle, up to 1, down to -1, and back.
So, within a period, they go twice through any value other than the extremes, -1 and 1, once on the way up, and another time on the way down, and in different quadrants.
{{{cos(x)<0}}} in quadrants II ({{{pi/2<x<=pi}}}), and III ({{{pi<=x<3pi/2}}}, so there will be a solution for {{{cos(x)=-sqrt(3)/2}}} in each of those quadrants.
{{{cos(pi/6)=sqrt(3)/2}}} is one of 5 numbers that you will see in many problems about trigonometric function values.
{{{-sqrt(3)/2}}} is the opposite value, corresponding to the supplementary angle, {{{pi-pi/6=highlight(5pi/6)}}} and to {{{pi+pi/6=highlight(7pi/6)}}}.
{{{drawing(300,300,-1.2,1.2,-1.2,1.2,grid(0),
circle(0,0,1),arrow(0,0,1.17,0.675),arrow(0,0,-1.17,0.675),arrow(0,0,-1.17,-0.675),
circle(0.866,0.5,0.03),circle(-0.866,0.5,0.03),circle(-0.866,-0.5,0.03),
locate(0.4,0.28,pi/6),locate(0.92,0.55,A),
locate(-0.5,0.28,pi/6),locate(-1,0.55,B),
locate(-0.52,0.02,pi/6),locate(-1,-0.45,C)
)}}} Points {{{"A ("}}}{{{sqrt(3)/2}}}{{{","}}}{{{1/2}}}{{{")"}}}, {{{"B ("}}}{{{-sqrt(3)/2}}}{{{","}}}{{{1/2}}}{{{")"}}}, and {{{"C ("}}}{{{-sqrt(3)/2}}}{{{","}}}{{{-1/2}}}{{{")"}}} are circled
{{{cos(pi-pi/6)=cos(5pi/6)=-sqrt(3)/2}}}
{{{cos(pi+pi/6)=cos(7pi/6)=-sqrt(3)/2}}}
 
EXTRA:
The 5 trigonometric function values that you will often see:
{{{cos(0)=highlight(1)=sin(pi/2)}}}
{{{cos(pi/6)=highlight(sqrt(3)/2)=sin(pi/3)}}}
{{{cos(pi/4)=highlight(sqrt(2)/2)=sin(pi/4)}}}
{{{cos(pi/3)=highlight(1/2)=sin(pi/6)}}}
{{{cos(pi/2)=highlight(0)=sin(0)}}}