sin(4x)-sin(6x)=0

Question

Solve for 0<x<2pi

 

Answer 1

sin 4x - sin 6x = 0

 

2 sin 2x cos 2x - 3 sin 2x + 4 sin³ 2x

 

(2 cos 2x + 4 sin² 2x - 3) sin 2x = 0

 

-(4 cos² 2x - 2 cos 2x - 1) sin 2x = 0

 

Case 1: sin 2x = 0

 

2x = kπ ; k ∈ Z

 

x = kπ/2 ; k ∈ Z

 

Case 2: 4 cos² 2x - 2 cos 2x - 1 = 0

 

cos 2x = [-(-2)±√((-2)² - 4(4)(-1))]/(2(4)) = (1±√5) / 4

 

2x = 2kπ ± π/5 or 2kπ ± 3π/5 ; k ∈ Z

 

x = kπ ± π/10 or kπ ± 3π/10 ; k ∈ Z

 

So the full solution set is:

 

{kπ/2 , kπ ± π/10 , kπ ± 3π/10 | k ∈ Z}

 

In the 0 < x < 2π range, this means:

 

{π/10,3π/10,π/2,7π/10,9π/10,π,11π/10,13π/10,3π/2,17π/10,19π/10}