Question

Find the general solution for \[cos4x = cos2x\]

Answer 1

Hint: To find the general solution of \[cos4x = cos2x\], we use the formulas of \[cosx - cosy = - 2sin\dfrac{{x + y}}{2}sin\dfrac{{x - y}}{2}\;\]. And using \[sinx = siny\;\] general solution, which is \[x = n\pi \pm {\left( { - 1} \right)^n}y\;\], we get the desired result.

Complete step by step Answer:

Given \[cos4x = cos2x\]
\[ \Rightarrow cos4x - cos2x = 0\]
As, we know that, \[cosx - cosy = - 2sin\dfrac{{x + y}}{2}sin\dfrac{{x - y}}{2}\;\]
Replacing x with \[4x\] and y with \[2x\], we get,
\[ \Rightarrow cos4x - cos2x = - 2sin\dfrac{{4x + 2x}}{2}sin\dfrac{{4x - 2x}}{2}\;\]
On simplification we get,
\[ \Rightarrow cos4x - cos2x = - 2sin3x\sin x\;\]
As, \[cos4x - cos2x = 0\], we get,
\[ \Rightarrow - 2sin3x\sin x = 0\;\]
\[ \Rightarrow sin3x\sin x = 0\;\]
So, either \[sin3x = 0\;\]or \[sinx = 0\]
We solve \[sin3x = 0\;\]& \[sinx = 0\] separately,
General solution for \[sin3x = 0\]
Let \[sinx = siny\;\]___(1)
\[sin3x = sin3y\;\]___(2)
Given \[sin3x = 0\;\]
From (1) and (2)
\[sin3y = 0\]
As, \[\sin (0) = 0\], we get,
\[ \Rightarrow sin3y = sin\left( 0 \right)\]
Using, if \[\sin x = \sin y\], then \[x = y\],
\[ \Rightarrow 3y = 0\]
\[ \Rightarrow y = 0\]___(3)
General solution for \[sin3x = sin3y\;\]is
\[3x = n\pi \pm {\left( { - 1} \right)^n}3y\;\]where \[n \in Z\]
Putting \[y = 0\], we get,
\[ \Rightarrow 3x = n\pi \pm {\left( { - 1} \right)^n}0\]
\[ \Rightarrow 3x = n\pi \]
On dividing by 3 we get,
\[ \Rightarrow x = \dfrac{{n\pi }}{3}\]​ where \[n \in Z\]
General solution for \[sinx = 0\]
Let \[sinx = siny\]
Given \[sinx = 0\],
From (1) and (2)
\[siny = 0\]
As, \[\sin (0) = 0\], we get,
\[ \Rightarrow siny = sin\left( 0 \right)\]
Using, if \[\sin x = \sin y\], then \[x = y\],
\[ \Rightarrow y = 0\]
General solution for \[sinx = siny\]is
\[x = n\pi \pm {\left( { - 1} \right)^n}y\;\]where \[n \in Z\]
putting \[y = 0\],
\[x = n\pi \pm {\left( { - 1} \right)^n}0\]
\[ \Rightarrow x = n\pi \]where \[n \in Z\]
Therefore,
General Solution are
For \[sin3x = 0,x = \dfrac{{n\pi }}{3}\]
Or
For \[sinx = 0,x = n\pi \]
Where \[n \in Z\]

Note: We need to keep in mind that \[sinx = siny\] has infinite number of solutions. For every n in this solution we will have different solutions in here \[x = n\pi \pm {\left( { - 1} \right)^n}y\;\]. You should consider both the cases, and solve for both of them.