Question 1137706


{{{ 2cos(2x)-2cosx=0}}} ,[{{{0}}},{{{2pi}}}) => means {{{0<=x<2pi}}}


{{{ 2cos(2x)-2cosx=0}}}

{{{2(cos(2x)-cosx)=0}}}


use identity

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

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

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

{{{2(2u ^2-u -1)=0}}}.....will be equal to zero if

{{{(2u ^2-u -1)=0}}} ..........use quadratic formula




{{{u=(-b +- sqrt(b^2-4ac))/2a}}}

{{{u=(-(-1)+- sqrt((-1)^2-4*2*(-1)))/(2*2)}}}

{{{u=(1+-sqrt(1+8))/4}}}

{{{u=(1+-sqrt(9))/4}}}

{{{u=(1+-3)/4}}}

solutions:

{{{u=4/4=1}}}

{{{u=-2/4=-1/2}}}


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

{{{cos(x)=1}}} => since given interval {{{0<=x<2pi}}} ,=> {{{x=0}}}

or {{{cos(x)=-1/2}}}=>{{{x = 2/3 (3pi* n + pi)}}}  or {{{x = 2/3 (3pi* n - pi)}}}, {{{n}}} element{{{ Z}}}

since given {{{0<=x<2pi}}} =>{{{x=2pi/3}}}, {{{x=4pi/3}}}

so, your solutions are: {{{x=0}}}, {{{x=2pi/3}}}, and {{{x=4pi/3}}}