2. \( \frac{2 \sin ^{2} x}{2 \tan x-\sin 2 x}=\frac{1}{\tan x} \)

To solve the equation \( \frac{2 \sin ^{2} x}{2 \tan x-\sin 2 x}=\frac{1}{\tan x} \), we begin by rewriting the terms in terms of sine and cosine. Recall that: - \( \tan x = \frac{\sin x}{\cos x} \) - \( \sin 2x = 2 \sin x \cos x \) Now, substituting these into the equation gives: \[ \frac{2 \sin^2 x}{2 \frac{\sin x}{\cos x} - 2 \sin x \cos x} = \frac{1}{\frac{\sin x}{\cos x}} \] Simplifying the left side: \[ \frac{2 \sin^2 x}{\frac{2 \sin x}{\cos x} - 2 \sin x \cos x} = \frac{1}{\frac{\sin x}{\cos x}} \] This reduces to: \[ \frac{2 \sin^2 x \cos x}{2 \sin x - 2 \sin x \cos^2 x} = \frac{\cos x}{\sin x} \] Cross-multiplying gives: \[ 2 \sin^2 x \cos^2 x = \cos x (2 \sin x - 2 \sin x \cos^2 x) \] Dividing both sides by \(\cos x\) (assuming \(\cos x \neq 0\)) yields: \[ 2 \sin^2 x \cos x = 2 \sin x - 2 \sin x \cos^2 x \] Rearranging gives us: \[ 2 \sin^2 x \cos x + 2 \sin x \cos^2 x = 2 \sin x \] Factoring out \(2 \sin x\): \[ 2 \sin x (\sin x \cos x + \cos^2 x) = 2 \sin x \] Assuming \(\sin x \neq 0\), we can divide both sides by \(2 \sin x\): \[ \sin x \cos x + \cos^2 x = 1 \] Here, we can use the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\) to rewrite the equation: \[ \sin x \cos x + (1 - \sin^2 x) = 1 \] This simplifies to: \[ \sin x \cos x = \sin^2 x \] Dividing both sides by \(\sin x\) (assuming \(\sin x \neq 0\)), we find: \[ \cos x = \sin x \] This results in: \[ \tan x = 1 \] The solutions for this equation are: \[ x = \frac{\pi}{4} + k\pi \quad \text{for } k \in \mathbb{Z} \] Thus, the final answer for the given equation is: \[ x = \frac{\pi}{4} + k\pi \quad (k \in \mathbb{Z}) \]