SOLUTION: 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

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

Use the general formula of Trigonometry

 = .                (2)

You have 

cos(x) + cos(4x) =  = ,

cos(2x) + cos(3x) =  = .

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

 = 0.                                (4)

Equation (4) deploys in two independent separate equations:


1.  cos(x) = 0  --->  x = ,  k = 0, +/-1, +/-2, . . . 


2.  cos(x/2) = 0  --->  x = ,  k = 0, +/-1, +/-2, . . . 


So, in the given interval the original equation has the roots 0, , ,   or  0°,  90°,  180°,  270°.


But these are not ALL the roots.

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 = ,  k = 0, +/-1, +/-2, . . . 

or, which is the same,

 =  + ,  k = 0, +/-1, +/-2, . . . 

So, these additional solutions are x = , , ,   PLUS  , k = 0, +/-1, +/-2, . . . 

The final answer is:  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°.