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

To solve the equation \( \frac{1 + 2 \sin x \cos x}{\sin x + \cos x} = \sin x + \cos x \), let's start by multiplying both sides by \( \sin x + \cos x \), assuming \( \sin x + \cos x \neq 0 \): \[ 1 + 2 \sin x \cos x = (\sin x + \cos x)(\sin x + \cos x) \] Expanding the right side, we have: \[ 1 + 2 \sin x \cos x = \sin^2 x + 2 \sin x \cos x + \cos^2 x \] Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \), we can simplify the expression: \[ 1 + 2 \sin x \cos x = 1 + 2 \sin x \cos x \] This simplifies to: \[ 1 + 2 \sin x \cos x = 1 + 2 \sin x \cos x \] This equation is tautological, meaning it holds true for all \( x \) where \( \sin x + \cos x \neq 0 \). Now, we need to find values of \( x \) where \( \sin x + \cos x = 0 \): \[ \sin x + \cos x = 0 \implies \sin x = -\cos x \] This occurs at: \[ \tan x = -1 \implies x = \frac{3\pi}{4} + n\pi, \quad n \in \mathbb{Z} \] Hence, the solution to the equation includes all \( x \) except for those where \( \sin x + \cos x = 0 \): \[ x = \frac{3\pi}{4} + n\pi, \quad n \in \mathbb{Z} \]