Solve the trigonometric equation: $\frac{\cos^3 x - \cos 3x}{\cos x} + \frac{\sin^3 x + \sin 3x}{\sin x} = 3$

1. Problem Description

Solve the trigonometric equation:

$\frac{\cos^3 x - \cos 3x}{\cos x} + \frac{\sin^3 x + \sin 3x}{\sin x} = 3$

2. Solution Steps

We use the triple angle formulas:

$\cos 3x = 4\cos^3 x - 3\cos x$

$\sin 3x = 3\sin x - 4\sin^3 x$

Substitute the triple angle formulas into the equation:

$\frac{\cos^3 x - (4\cos^3 x - 3\cos x)}{\cos x} + \frac{\sin^3 x + (3\sin x - 4\sin^3 x)}{\sin x} = 3$

Simplify the expression:

$\frac{\cos^3 x - 4\cos^3 x + 3\cos x}{\cos x} + \frac{\sin^3 x + 3\sin x - 4\sin^3 x}{\sin x} = 3$

$\frac{-3\cos^3 x + 3\cos x}{\cos x} + \frac{-3\sin^3 x + 3\sin x}{\sin x} = 3$

Assuming $\cos x \neq 0$ and $\sin x \neq 0$, we can divide by $\cos x$ and $\sin x$ respectively:

$-3\cos^2 x + 3 + (-3\sin^2 x + 3) = 3$

$-3\cos^2 x + 3 - 3\sin^2 x + 3 = 3$

$-3(\cos^2 x + \sin^2 x) + 6 = 3$

Since $\cos^2 x + \sin^2 x = 1$:

$-3(1) + 6 = 3$

$-3 + 6 = 3$

$3 = 3$

This means that the equation is an identity, and it holds true for all $x$ such that $\cos x \neq 0$ and $\sin x \neq 0$.

Therefore, $x \neq k\frac{\pi}{2}$ for $k \in \mathbb{Z}$ (integers).

$x$ can be any real number except integer multiples of $\frac{\pi}{2}$.

3. Final Answer

$x \neq k\frac{\pi}{2}$, where $k$ is an integer.