Question 1138018
{{{cos(2x) = -cos(x)}}} for the interval {{{0 <= x < 2pi}}}

 {{{cos (2x)+cos (x) =0}}}..........use the following identity : {{{cos  (2x )= -1+2 cos^2 (x)}}}

{{{(-1+2cos^2 (x))+cos (x) =0}}}

{{{2cos^2 (x) +cos (x)-1 =0}}}.......let {{{cos (x) =u}}}

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

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

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

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

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


solutions:

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

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


since {{{cos (x) =u}}}, we have

{{{cos (x) =-1}}} or

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


if {{{cos (x )=-1}}}, and {{{0<=x<2pi}}}, then {{{x=pi }}}

if {{{cos (x )=1/2}}}, and{{{ 0<=x<2pi}}}, than {{{x=pi/3}}}, {{{x=5pi/3 }}}


combine all solutions: {{{x=pi/3}}}, {{{x=5pi/3 }}},{{{x=pi }}}