Question 1063391
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Solve the following equations for 0deg <= x <= 360deg:
(i) cos 2x cos x = sin 4x sin x
(ii) cos x + cos 2x + cos 3x + cos 4x = 0
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I will solve (ii).


<pre>
cos(x) + cos(2x) + cos(3x) + cos(4x) = 0.        (1)

Use the general formula of Trigonometry

{{{cos(a) + cos(b)}}} = {{{2cos((a+b)/2)*cos((a-b)/2)}}}.                (2)

You have 

cos(x) + cos(4x) = {{{2*cos((x+4x)/2)*cos((4x-x)/2)}}} = {{{2*cos(2.5x)*cos(1.5x)}}},

cos(2x) + cos(3x) = {{{2*cos((2x+3x)/2)*cos((3x-2x)/2)}}} = {{{2*cos(2.5x)*cos(0.5x)}}}.

Therefore, the left side of the original equation is

cos(x) + cos(2x) + cos(3x) + cos(4x) = 2*cos(2.5x)*cos(1.5x) + 2*cos(2.5x)*cos(0.5x) = 2*cos(2.5x)*(cos(1.5x) + cos(0.5x)).

Hence, the original equation is equivalent to

2*cos(2.5x)*(cos(1.5x) + cos(0.5x)) = 0,    or,  canceling  the factor  2*cos(2.5x),

cos(1.5x) + cos(0.5x) = 0.                       (3)

Again, apply the formula (2) to the left side of (3). You will get an equivalent equation

{{{2*cos(x)*cos(x/2)}}} = 0.                                (4)

Equation (4) deploys in two independent separate equations:


1.  cos(x) = 0  --->  x = {{{pi/2 + k*pi}}},  k = 0, +/-1, +/-2, . . . 


2.  cos(x/2) = 0  --->  x = {{{k*pi}}},  k = 0, +/-1, +/-2, . . . 


So, in the given interval the original equation has the roots 0, {{{pi/2}}}, {{{pi}}}, {{{3pi/2}}}  or  0°,  90°,  180°,  270°.


<U>But these are not ALL the roots</U>.

There is one more family of roots.

Do you remember I canceled the factor 2*cos(2.5x) ?

Of course, I must consider (and add !) all the solutions of the equation

cos(2.5x) = 0.

They are  2.5x = {{{pi/2 + k*pi}}},  k = 0, +/-1, +/-2, . . . 

or, which is the same,

{{{(5x)/2}}} = {{{pi/2}}} + {{{k*pi}}},  k = 0, +/-1, +/-2, . . . 

So, these additional solutions are x = {{{pi/5}}}, {{{2pi/5}}}, {{{3pi/5}}}, {{{4pi/5}}}  PLUS  {{{k*pi}}}, k = 0, +/-1, +/-2, . . . 

<U>The final answer is</U>:  There are two families of solutions. 

                      One family is 0°, 90°, 180° and 270°.

                      The other family is 36°, 72°, 108°, 144°, 180°, 216°, 252°, 288°, 324°.
</pre>

Solved.