Question 960547
.
Find all solutions in 0 ≤ x < 2π.
Use exact values only. (Enter your answers as a comma-separated list.)
cos 2x cos x − sin 2x sin x = square root of 2 over 2
x=?
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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;In his post, @lwsshar3 found two solutions, &nbsp;{{{pi/12}}} &nbsp;and &nbsp;{{{7pi/12}}}.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;Actually, &nbsp;this equation has &nbsp;6 &nbsp;(six) &nbsp;solutions, &nbsp;but @lwsshar3 missed/lost most of them.

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;See my correct and complete solution below.



<pre>
Use identity:  cos(x+y) = cos(x)*cos(y) - sin(x)*sin(y)

cos(2x)* cos(x) − sin(2x)* sin( x) = square root of 2 over 2

cos(2x+x) = √2/2

2x+x = π/4 + 2k*π;     7π/4 + 2k*π.

  3x = π/4 + 2k*π;     7π/4 + 2k*π.


    For 3x, you must add the periods 2π and 4π for cosine.
    For 3x, it will produce geometrically the same angle; 
    but for 'x' it will produce new angles that you will miss otherwise.

    You should no add more hire periods for 3x, since it will not produce 
    new angles for x and will lead you out of the given interval.


Now consider two cases.


Case (a).  3x = {{{pi/4}}},  {{{pi/4+2pi}}},  {{{pi/4+4pi}}}.

            then  x = {{{pi/12}}},  {{{pi/12+(2pi)/3}}} = {{{9pi/12}}} = {{{(3/4)pi}}},  {{{pi/12+(4pi)/3}}} = {{{17pi/12}}}.



Case (b).  3x = {{{7pi/4}}},  {{{7pi/4+2pi}}},  {{{7pi/4+4pi}}}.

            then  x = {{{7pi/12}}},  {{{7pi/12+(2pi)/3}}} = {{{15pi/12}}} = {{{(5/4)pi}}},  {{{7pi/12+(4pi)/3}}} = {{{23pi/12}}}.



<U>ANSWER</U>.  The given equation has 6 solutions in the given interval

         {{{pi/12}}},  {{{(3/4)pi}}},  {{{17pi/12}}},  {{{7pi/12}}},  {{{(5/4)pi}}},  {{{23pi/12}}}.
</pre>

Solved completely and correctly - no one root is missed.


This analysis in my post is typical in trigonometry problems, when you work with a multiple of an angle.


If you miss this analysis, you will miss many solutions to your original equation.