Question 1161422


{{{12cos^2 (x)-4cos(x) =1}}}      {{{0 <=x <2pi}}}

let {{{cos (x )=u}}}

{{{12u^2-4u =1  }}}

{{{12u^2-4u -1  =0}}}......factor

{{{12u^2+2u -6u-1  =0}}}

{{{(12u^2+2u )-(6u+1 ) =0}}}

{{{2u(6u+1 )-(6u+1 ) =0}}}

{{{(2u - 1) (6u + 1) = 0}}}


solutions:

if {{{(2u - 1)  = 0}}} ->{{{2u=1}}}->{{{u=1/2}}}

if {{{(6u + 1) = 0}}}->{{{6u=-1}}}->{{{u=-1/6}}}


substitute back u

{{{cos (x )=1/2}}}

{{{cos (x )=-1/6}}}


{{{x=cos^-1 (1/2)+2pi*n }}} ->{{{x=pi/3+2pi*n}}}

{{{x=cos^-1 (-1/6) +2pi*n}}} 


since given {{{0 <= x < 2pi}}}, use solutions

{{{highlight(x=pi/3)}}} and {{{highlight(x=5pi/3)}}}