We are asked to simplify the expression: $\frac{\sin x + \sin 2x + \sin 3x}{\cos x + \cos 2x + \cos 3x}$ and then compare it to $\sqrt{3}$

1. Problem Description

We are asked to simplify the expression:

$\frac{\sin x + \sin 2x + \sin 3x}{\cos x + \cos 2x + \cos 3x}$

and then compare it to $\sqrt{3}$

2. Solution Steps

We can rewrite the expression using sum-to-product formulas.

$\sin A + \sin B = 2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})$

$\cos A + \cos B = 2 \cos(\frac{A+B}{2}) \cos(\frac{A-B}{2})$

First, we group $\sin x$ and $\sin 3x$ in the numerator:

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

(Since $\cos(-x) = \cos(x)$)

So, the numerator becomes:

$\sin x + \sin 2x + \sin 3x = 2 \sin(2x) \cos x + \sin 2x = \sin 2x (2 \cos x + 1)$

Similarly, we group $\cos x$ and $\cos 3x$ in the denominator:

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

So, the denominator becomes:

$\cos x + \cos 2x + \cos 3x = 2 \cos(2x) \cos x + \cos 2x = \cos 2x (2 \cos x + 1)$

Therefore, the expression becomes:

$\frac{\sin x + \sin 2x + \sin 3x}{\cos x + \cos 2x + \cos 3x} = \frac{\sin 2x (2 \cos x + 1)}{\cos 2x (2 \cos x + 1)}$

If $2 \cos x + 1 \ne 0$, we can cancel the common factor:

$\frac{\sin 2x (2 \cos x + 1)}{\cos 2x (2 \cos x + 1)} = \frac{\sin 2x}{\cos 2x} = \tan 2x$

We are given that $\tan 2x = \sqrt{3}$.

The general solution for $\tan \theta = \sqrt{3}$ is $\theta = n\pi + \frac{\pi}{3}$, where $n$ is an integer.

Therefore, $2x = n\pi + \frac{\pi}{3}$, so $x = \frac{n\pi}{2} + \frac{\pi}{6}$, where $n$ is an integer.

3. Final Answer

$\tan 2x = \sqrt{3}$

$x = \frac{n\pi}{2} + \frac{\pi}{6}$, where n is an integer.