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

To solve the equation \( \frac{\sin x+\sin 2 x}{1+\cos x+\cos 2 x}=\tan x \), let's first rewrite both sides in a more manageable form. The left-hand side can be expressed using the double angle identity for sine: \[ \sin 2x = 2 \sin x \cos x. \] Thus, the left side becomes: \[ \frac{\sin x + 2 \sin x \cos x}{1 + \cos x + \cos 2x}. \] Now, we know that \( \cos 2x = 2\cos^2 x - 1 \), so we substitute that into the denominator: \[ 1 + \cos x + (2 \cos^2 x - 1) = \cos x + 2 \cos^2 x. \] So now the left-hand side can be represented as: \[ \frac{\sin x(1 + 2 \cos x)}{\cos x + 2 \cos^2 x}. \] For the right-hand side, \( \tan x \) is defined as \( \frac{\sin x}{\cos x} \). Thus, the equation becomes: \[ \frac{\sin x(1 + 2 \cos x)}{\cos x + 2 \cos^2 x} = \frac{\sin x}{\cos x}. \] Assuming \(\sin x \neq 0\) (we will handle that case separately), we can divide both sides by \(\sin x\): \[ \frac{1 + 2 \cos x}{\cos x + 2 \cos^2 x} = \frac{1}{\cos x}. \] Cross-multiplying gives us: \[ (1 + 2 \cos x) \cos x = \cos x + 2 \cos^2 x. \] Expanding both sides results in: \[ \cos x + 2 \cos^2 x = \cos x + 2 \cos^2 x. \] This is always true, suggesting that all \(x\) satisfying \(\sin x \neq 0\) are valid solutions. Now, let's check the case when \(\sin x = 0\), which occurs at \(x = n\pi\) for \(n\) an integer. For \(x = n\pi\): \[ \tan(n\pi) = 0, \] and the left-hand side: \[ \frac{\sin(n\pi) + \sin(2n\pi)}{1 + \cos(n\pi) + \cos(2n\pi)} = \frac{0}{1 + (-1)^n + 1} = 0. \] Hence, the equation holds for \(x = n\pi\) as well. In conclusion, the solutions to the equation are: \[ x = n\pi \quad (n \in \mathbb{Z}). \]