Question 1116853
1. An  exact general solution {{{highlight(x=(12n+8 +- 1)pi/18)}}} .
2. The particular solutions in {{{"[ 0 ,"}}}{{{2pi}}}{{{")"}}}
can found by using the formula above for {{{"n = 0 , 1 , 2"}}}
{{{x=highlight(7pi/18)}}} and {{{x=9pi/18=highlight(pi/2)}}} for {{{n=0}}} ,
{{{x=highlight(19pi/18)}}} and {{{x=21pi/18=highlight(7pi/6)}}} for {{{n=1}}} , and
{{{x=highlight(31pi/18)}}} and {{{x=33pi/18=highlight(11pi/6)}}} for {{{n=2}}} .


EXPLANATION AND GRAPHICAL REORESENTATION:
As Edwin showed you, simple algebra leads you to
{{{sin(3x+ pi/6)=-sqrt(3)/2}}} .
 
What angles could measure {{{3x+ pi/6}}} ?
{{{drawing(300,300,-1.2,1.2,-1.2,1.2,grid(0),
circle(0,0,1),triangle(-0.5,-0.866,0.5,-0.866,0,0),
arrow(0,0,-0.65,-1.126),arrow(0,0,0.65,-1.126),
arrow(0,0,-1.039,-0.6),arrow(0,0,1.039,-0.6),
locate(1.02,0.12,A),locate(-0.52,-0.88,B),
locate(0.53,-0.83,D),locate(0.02,-1,C),
locate(0.02,0.12,O),red(arc(0,0,0.9,0.9,0,30)),
red(arc(0,0,1,1,30,60)),red(arc(0,0,0.9,0.9,60,90)),
red(arc(0,0,1,1,90,120)),red(arc(0,0,0.9,0.9,120,150)),
red(arc(0,0,1,1,150,180)),locate(0.38,-0.3,red(pi/6))
)}}}
The angles {{{AOB}}} and {{{AOD}}} have
{{{sin(AOB)=sin(AOD)=-sqrt(3)/2=about-0.866}}} .
One whole counterclockwise turn is {{{2pi}}} (radians),
{{{AOC=3pi/2}}} .
Angles {{{AOB}}} and {{{AOD}}} measure {{{pi/6}}} less, and {{{pi/6}}} more than that,
or {{{4pi/3}}} and {{{5pi/3}}} respectively
A general formula for clockwise expression for the measure of {{{AOB}}} or {{{AOD}}} is
{{{3pi/2 +- pi/6=(9 +- 1)pi/6}}} ,
and a general formula for all angles co-terminal with those is
{{{(9 +- 1)pi/6 +2n*pi=(12n+9 +- 1)pi/6}}} for any integer {{{n}}} .
So, to find the exact general solution, with {{{n}}} representing any integer
{{{3x+ pi/6=(12n+9 +- 1)pi/6}}}
{{{3x=(12n+8 +- 1)pi/6 - pi/6}}}
{{{3x=(12n+8 +- 1)pi/6}}}
{{{x=(12n+8 +- 1)pi/18}}}


RANT:
In math, there is no such a thing as "this kind of question"
or "this kind of problem".
Unfortunately, too many math instructors and tutors
classify problems into hundreds or thousands of "kinds" or "types",
and then encourage students to memorize the classification,
and the corresponding problem-solving recipes.
That makes math into a no-reasoning memorization exercise similar to learning a foreign language.
Why?
Maybe because is easier than trying to persuade students that they only need to
1) understand a few simple concepts, and
2) apply those concepts and their own brains.
Maybe because they were taught math that same way,
and they do not know any better.
Or is it that they believe the students are incapable of reasoning?