Solve: CosxCos4x=Cos2xCos3x

Question:

Cosx Cos4x=Cos2x Cos3x

Solution:

Using , Cos(a)Cos(b) = $\frac{Cos(a-b)+Cos(a+b)}{2}$

We would have,

Cos(4x)Cos(x) = $\frac{Cos(4x-x)+Cos(4x+x)}{2}$

Cos(2x)Cos(3x) = $\frac{Cos(2x-3x)+Cos(2x+3x)}{2}$

hence, 

$\frac{Cos(3x)+Cos(5x)}{2}$ = $\frac{Cos(-x)+Cos(5x)}{2}$        [Recall , Cos(-a)= Cos(a)]

             $Cos(3x)Cos(5x) = Cos(x)Cos(5x)$

             $Cos(3x) = Cos(x)$

So, (3x+$2n_{1}\pi$ )= (-x + $2n_{2}\pi$)     [Recall, arccos(Cos(x))= (x+$2n\pi$) or (-x+$2n\pi)$]

              4x =  2($n_{1}$- $n_{2}$).($\pi$)

              $4x = 2n\pi$                        [$n_{1}$- $n_{2}$ = n]

              $x = \frac{n\pi}{2}$