Question 675749
replace {{{cos(2x)}}} with {{{1-2*sin^2(x)}}}

{{{1-2sin^2(x)=2/3}}} 

{{{-2sin^2(x)=2/3-1}}} 

{{{-2sin^2(x)=-1/3}}} 

{{{2sin^2(x)=1/3}}}
 
{{{sin^2(x)=1/6}}} 

{{{sin(x)= sqrt(1/6)}}} 

since {{{sin(x)}}} is negative in quad III, solution is {{{sin(x)= -sqrt(1/6)=-0.408}}}

{{{x=-24.08}}} degrees

or
{{{x=180-(-24.08)=180+24.08=204.08}}} degrees.....in quadrant III