Question

Find the general solution for $\cos 4x=\cos 2x$

• A. $x=n\pi$ or $\frac{n\pi }{6}$
• B. $x=n\pi$ or $\frac{n\pi }{3}$
• C. $x=\frac{2n\pi }{3}$
  
• D. $x=\pi$
  

Answer 1

The correct option is B $x=n\pi$ or $\frac{n\pi }{3}$

$\cos 4x=\cos 2x$

$\cos 4x-\cos 2x=0$

we know that $\cos x-\cos y=-2\sin \frac{x+y}{2}\sin \frac{x-y}{2}$

Replacing x with $4x$ and y with $2x$

$-2\sin \left(\frac{4x+2x}{2}\right)\sin \left(\frac{4x-2x}{2}\right)=0$

$-2\sin \left(\frac{6x}{2}\right)\sin \left(\frac{2x}{2}\right)=0$

$-2\sin 3x\sin x=0$

$\sin 3x\sin x=\frac{0}{-2}$

$\sin 3x\sin x=0$

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)

$\sin 3y=0$

$\sin 3y=\sin \left(0\right)$

$3y=0$ ___(3)

$y=0$ ___(4)

General solution for $\sin 3x=\sin 3y$ is

$3x=n\pi \pm (-1{)}^{n}3y$ where $n\in Z$

Putting $y=0$

$3x=n\pi \pm (-1{)}^{n}0$

$3x=n\pi$

$x=\frac{n\pi }{3}$ where $n\in Z$

General solution for $\sin x=0$

Let $\sin x=\sin y$

Given $\sin x=0$

From (1) and (2)

$\sin y=0$

$\sin y=\sin \left(0\right)$

$y=0$

General solution for $\sin x=\sin y$ is

$x=n\pi \pm (-1{)}^{n}y$ where $n\in Z$

putting $y=0$

$x=n\pi \pm (-1{)}^{n}0$

$x=n\pi$ where $n\in Z$

Therefore,

General Solution are

For $\sin 3x=0,x=\frac{n\pi }{3}$

Or

For $\sin x=0,x=n\pi$

Where $n\in Z$