General solution of equation sin x + sin 5x = sin 2x + sin 4x is

x=nπ3, n∈I

Step-by-step explanation:

sin x + sin 5x = sin 2x + sin 4x

⇒⇒   2 sin 3x cos 2x = 2 sin 3x cos x

⇒⇒   2 sin 3x(cos 2x - cos x) = 0

⇒    sin 3x=0 ⇒ 3x=nπ ⇒ x=nπ3⇒    sin 3x=0 ⇒ 3x=nπ ⇒ x=nπ3 , n∈I, n∈I

or    cos 2x−cos x=0 ⇒ cos 2x=cos xcos 2x-cos x=0 ⇒ cos 2x=cos x

⇒   2x=2nπ±x  ⇒  x=2nπ, 2nπ3⇒   2x=2nπ±x  ⇒  x=2nπ, 2nπ3 , n∈I, n∈I

But solutions obtained by x=2nπx=2nπ , n∈I, n∈I or x=2nπ3x=2nπ3 , n∈I, n∈I are all involved in x=nπ3x=nπ3 , n∈I

sorry not sure

step-by-step explanation: