Question 1152146


Find all solutions of the equation {{{cos(3x)=cos(x)}}} in the interval [{{{0}}},{{{2pi}}})

{{{cos(3x)=cos(x)}}}......write {{{cos(3x)}}} as {{{cos(2x+x)}}} and use identity {{{cos(alpha + beta) = cos(alpha) cos(beta) - sin(alpha) sin(beta)}}}


{{{cos(2x+x)= cos(2x) cos(x)  -  sin(2x) sin(x)}}}

now, write {{{cos(2x)}}} as {{{cos(x+x)}}} and use identity above

{{{cos(x+x)= cos(x) cos(x)  -  sin(x) sin(x)}}}


and {{{sin(2x)=sin(x+x)=sin(x) cos(x) + cos(x) sin(x)}}}


then we have

{{{cos(3x)=(cos(x) cos(x) -sin(x) sin(x)) cos(x) - (sin(x) cos(x) + cos(x) sin(x)) sin(x)}}}


substitute in given equation:

{{{cos(3x)=cos(x)}}}

{{{(cos(x) cos(x) -sin(x) sin(x)) cos(x) - (sin(x) cos(x) + cos(x) sin(x)) sin(x)=cos(x)}}}

{{{(cos^2(x)  -sin^2(x) ) cos(x) - (sin^2(x) cos(x) + cos(x) sin^2(x)) =cos(x)}}}

{{{(cos^2(x)  -sin^2(x) ) cos(x) - (sin^2(x)  + sin^2(x)) cos(x) =cos(x)}}}....both sides divide by  {{{cos(x)}}}

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

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

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

{{{  1-cos^2(x)+3sin^2(x)=0}}}........use the identity: {{{1-cos ^2(x)=sin ^2(x)}}}

{{{sin ^2(x)+3sin^2(x)=0}}}

{{{4sin^2(x)=0}}}

=> solutions:


if {{{sin^2(x)=0}}} ...........apply rule: {{{x^n=0}}}=>{{{x=0}}}

 general solutions for {{{sin(x)=0}}} are:

{{{x=0+2pi*n}}} =>{{{ x=2pi*n}}}
{{{x=pi+2pi*n}}}

solutions for the range : {{{0<=x<2pi }}}

{{{x=0}}}, {{{x=pi}}}

combine solutions:

{{{x=pi/2}}},{{{ x=0}}}, {{{x=3pi/2}}},{{{x=pi}}}