Question 1132562
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1 + cos(6x) - 2cos(4x) = 0


The idea of solution is to express cos(6x) via cos(2x) and express cos(4x) via cos(2x) to get in this way

an equation for cos(2x) uniformly in the left side.



For it, use widely known formulas of Trigonometry

    cos(3a) = {{{4*cos^3(a) - 3*cos(a)}}},      (1)   and

    cos(2a) = {{{2*cos^2(a) - 1}}}.           (2)



Use a = 2x;  then formulas (1) and (2) give you

    cos(6x) = {{{4*cos^3(2x) - 3*cos(2x)}}},     (3)   and

    cos(4x) = {{{2*cos^2(2x) - 1}}}.           (4)



Now substitute (3) and (4) into your basic equation. You will get

    {{{1 + 4*cos^3(2x) - 3*cos(2x) - 4*cos^2(2x) + 2}}} = 0,    or

    {{{4*cos^3(2x) - 4*cos^2(2x) - 3*cos(2x) + 3}}} = 0.    (5)



In the last equation, introduce new variable  y = cos(2x).  You will get then

    {{{4y^3 - 4y^2 - 3y + 3}}} = 0.


You can group the terms and factor it in this way

    {{{4y^2*(y-1) - 3*(y-1)}}} = 0,

    {{{(y-1)*(4y^2-3)}}} = 0.


So the roots for y are  {{{y[1]}}} = 1;  {{{y[2]}}} = {{{sqrt(3)/2}}}  and  {{{y[3]}}} = -{{{sqrt(3)/2}}}.



Thus we have three cases to analyse separately:


    <U>Case 1</U>.  {{{y[1]}}} = 1  ====>  cos(2x) = 1  ====>  2x = 0  or  {{{2*pi}}}  ====>  x = 0  or  x = {{{pi}}}.  


    <U>Case 2</U>.  {{{y[2]}}} = {{{sqrt(3)/2}}}  ====>  cos(2x) = {{{sqrt(3)/2}}}  ====>  2x = {{{pi/6}}}  or  2x = {{{11pi/6}}}  ====>  x = {{{pi/12}}}  or  x = {{{11pi/12}}}. 


    <U>Case 3</U>.  {{{y[3]}}} = -{{{sqrt(3)/2}}}  ====>  cos(2x) = -{{{sqrt(3)/2}}}  ====>  2x = {{{5pi/6}}}  or  2x = {{{7pi/6}}}  ====>  x = {{{5pi/12}}}  or  x = {{{7pi/12}}}. 



<U>ANSWER</U>.  In all, there are 6 solutions for x in the given interval:

         x = 0;  {{{pi/12}}};  {{{5pi/12}}};  {{{7pi/12}}};  {{{11pi/12}}}  and  {{{pi}}}.
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Solved.