Guys please help me with this Trigonometry question based on De Moivre's theorem.

Question

Q: Find ¦È such that 0¡Ü¦È¡Ü360.
cos7¦È + cos3¦È = cos5¦È

I attempted to solve it with limited knowledge but I kind of doubt my answer. Anyway, this is how I did it :
1) [cos7¦È + cos3¦È] - cos5¦È = 0

2) (cos5¦È x cos3¦È) - cos5¦È = 0
[Application of cos (C) and cos (D)
= cos5(C+D) x cos5(C-D)]

3) cos5¦È[cos3¦È - 1] = 0

4) => either cos5¦È = 0 or cos3¦È = 1

5) So, when cos5¦È = 0, ¦È = (2n ¡À 1)(¦Ð/4) wher n is an integer and when cos¦È = 1
¦È=2n¦Ð where n is an integer.

I'm stuck here. Please show me step by step how to solve this. :/

Answer 1

I am going to rephrase your question, without fancy symbols

cos(7E) + cos(3E) = cos(5E) , where 0 ≤ E ≤ 360°

Correction on the formula you are using:
should say:
cosA + cosB = 2cos((A+B)/2)cos((A-B)/2)

so cos(7E) + cos(3E)
= 2cos(5E)cos(2E)
but cos(7E) + cos(2E) = cos(5E) <----- given

thus:
2cos(5E)cos(2E) = cos(5E)
2cos(5E)cos(2E) - cos(5E) = 0
cos 5E (2cos 2E - 1) = 0
cos 5E = 0 or cos 2E = 1/2

case1,
if cos 5E = 0
5E = 90° or 5E = 270°
E = 18° or E = 54°
The period of cos 5E is 360/5° = 72°
so adding multiples of 72 to each answer yields a new answer.
E = 18°, 90° 162°, 234°, 306°, 54°, 126°, 198°, 270°, 342°

case 2
cos 2E = 1/2
2E = 60° or 2E = 300°
E = 30° or E = 150°
to get the other angles, proceed in the same way I showed you for case1