Prove the following: \[ \frac{3 \sin x+2 \sin 2 x}{2+3 \cos x+2 \cos 2 x}=\tan x \]

To prove the identity \[ \frac{3 \sin x + 2 \sin 2x}{2 + 3 \cos x + 2 \cos 2x} = \tan x, \] we start by expressing everything in terms of \( \sin x \) and \( \cos x \). First, we know that \[ \sin 2x = 2 \sin x \cos x \] and \[ \cos 2x = 2 \cos^2 x - 1. \] Substituting these identities into the equation gives: \[ \text{Numerator: } 3 \sin x + 2 \sin 2x = 3 \sin x + 2(2 \sin x \cos x) = 3 \sin x + 4 \sin x \cos x = \sin x (3 + 4 \cos x). \] Now, for the denominator: \[ \text{Denominator: } 2 + 3 \cos x + 2 \cos 2x = 2 + 3 \cos x + 2(2 \cos^2 x - 1) = 2 + 3 \cos x + 4 \cos^2 x - 2 = 4 \cos^2 x + 3 \cos x. \] Now the entire expression looks like: \[ \frac{3 \sin x + 2 \sin 2x}{2 + 3 \cos x + 2 \cos 2x} = \frac{\sin x (3 + 4 \cos x)}{4 \cos^2 x + 3 \cos x}. \] Next, we simplify the denominator: \[ 4 \cos^2 x + 3 \cos x = \cos x (4 \cos x + 3). \] Rearranging the fraction now gives us: \[ \frac{\sin x (3 + 4 \cos x)}{\cos x (4 \cos x + 3)}. \] This can be separated into two fractions: \[ \frac{\sin x}{\cos x} \cdot \frac{3 + 4\cos x}{4 \cos x + 3}. \] Notice that: \[ \frac{\sin x}{\cos x} = \tan x. \] Now, we need to check if \(\frac{3 + 4 \cos x}{4 \cos x + 3} = 1\): Since both the numerator and denominator are the same: \[ 3 + 4\cos x = 4\cos x + 3. \] Thus, we have \[ \frac{3 + 4 \cos x}{4 \cos x + 3} = 1. \] Therefore, we arrive at: \[ \frac{\sin x (3 + 4 \cos x)}{4 \cos^2 x + 3 \cos x} = \frac{\sin x}{\cos x} = \tan x. \] Thus, we have proven that \[ \frac{3 \sin x + 2 \sin 2x}{2 + 3 \cos x + 2 \cos 2x} = \tan x. \] Hence, the identity holds true.