Question 871008
{{{2cos(3x)=1}}}
{{{cos(3x)=1/2}}}
The cosine function takes all the positive values between 0 and 1, without repeating in quadrant I and again in quadrant IV.
In quadrants II and III cosine has negative values.
In quadrant I (between 0 and {{{pi/2}}} , or if you prefer between{{{0^o}}} and {{{90^o}}} ),
we have {{{cos(pi/3)=1/2}}} .
In quadrant IV , {{{-pi/3}}} has a cosine of {{{1/2}}} too.
Since cosine has a period of {{{2pi}}} , adding {{{2pi}}} you get an angle with the same cosine.
So all the angles with a cosine of {{{1/2}}} can be expressed as
{{{2k*pi +- pi/3}}}={{{(6k +- 1)pi/3}}} , for any integer {{{k}}}
If {{{cos(3x)=1/2}}} , then {{{3x=(6k +- 1)pi/3}}} --> {{{x=(6k +- 1)pi/9}}} .
{{{(6*0 - 1)pi/9=-pi/9<0}}} and {{{(6*3+1)pi/9=highlight(19pi/9>2pi)}}} are outside the interval {{{"[ 0 ,"}}}{{{2pi}}}{{{")"}}} .
The solutions in the interval {{{"[ 0 ,"}}}{{{2pi}}}{{{")"}}} are
{{{(6*0 + 1)pi/9=highlight(pi/9)}}} , {{{(6*1 - 1)pi/9=highlight(5pi/9)}}} , {{{(6*1 + 1)pi/9=highlight(7pi/9)}}} , {{{(6*2 - 1)pi/9=highlight(11pi/9)}}} , {{{(6*2+1)pi/9=highlight(13pi/9)}}} , {{{(6*3 - 1)pi/9=highlight(17pi/9)}}} .