cos 4x = cos 2x
cos 4x - cos 2x = 0
We know that
cos x - cos y = -2sin \(\frac{x + y}{2} sin \frac{x - y}{2}\)
Replacing x with 4x and y with 2x
sin 3x sin x = \(\frac{0}{-2}\)
sin 3x sin x = 0
So, either sin 3x = 0 or sin x = 0
We solve sin 3x = 0 & sin x = 0 separately
General solution for sin 3x = 0
Let sin x = sin y ........(1)
sin 3x = sin 3y .........(2)
Given sin 3x = 0
From (1) and (2)
sin 3y = 0
sin 3y = sin (0) ......(3)
3y = 0 .......(4)
y = 0
General solution for sin 3x = sin 3y is
\(3x = n \pi \pm(-1)^n\, 3y\) where n ∈ z.
Putting y = 0
General solution for sin x = 0
Let sinx = sin y
Given sin x = 0
From (1) and (2)
sin y = 0
sin y = sin (0)
y = 0
General solution for sin x = sin y is
\(x = n \pi \pm(-1)^n\, y\) where n ∈ z.
Putting y = 0
\(x = n \pi \pm (-1)^n 0\)
\(x = n \pi\) where n ∈ z.
Therefore,
General Solution are
For sin 3x = 0, \(x = \frac{n \pi}{3}\)
Or
For sin x = 0, x = nπ, where n ∈ z.
