Question

Find the general solution for $cos4x = cos2x$

Answer 1

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.