2. $$ \frac{2 \sin ^{2} x}{2 an x-\sin 2 x}=\frac{1}{ an x} $$

Question

2. $$ \frac{2 \sin ^{2} x}{2 	an x-\sin 2 x}=\frac{1}{	an x} $$

UpStudy AI · Accepted Answer

We start with the equation

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

**Step 1.** Rewrite $$\sin 2x$$ using the double‑angle formula:

$$
\sin 2x = 2 \sin x \cos x.
$$

Then the denominator becomes

$$
2 \tan x - \sin 2x = 2 \tan x - 2 \sin x \cos x.
$$

**Step 2.** Express $$\tan x$$ as $$\frac{\sin x}{\cos x}$$ to have everything in terms of $$\sin x$$ and $$\cos x$$:

$$
2 \tan x - 2 \sin x \cos x = 2\left(\frac{\sin x}{\cos x}\right) - 2 \sin x \cos x.
$$

So the original equation becomes

$$
\frac{2 \sin^2 x}{2\left(\frac{\sin x}{\cos x}\right) - 2 \sin x \cos x} = \frac{1}{\tan x}.
$$

**Step 3.** Factor and simplify the left-hand side (LHS). First, factor $$2\sin x$$ from the denominator:

$$
2\left(\frac{\sin x}{\cos x}\right) - 2 \sin x \cos x = 2 \sin x\left(\frac{1}{\cos x} - \cos x\right).
$$

Thus, the LHS becomes

$$
\frac{2 \sin^2 x}{2 \sin x \left(\frac{1}{\cos x} - \cos x\right)} = \frac{\sin x}{\frac{1}{\cos x} - \cos x},
$$

provided that $$\sin x \neq 0$$.

On the right-hand side (RHS), since

$$
\tan x = \frac{\sin x}{\cos x},
$$

we have

$$
\frac{1}{\tan x} = \frac{\cos x}{\sin x},
$$

with the condition that $$\tan x$$ is defined (so $$\cos x \neq 0$$) and $$\sin x \neq 0$$ to avoid division by zero.

**Step 4.** The equation is now

$$
\frac{\sin x}{\frac{1}{\cos x} - \cos x} = \frac{\cos x}{\sin x}.
$$

Cross multiply (keeping in mind the restrictions):

$$
\sin x \cdot \sin x = \cos x\left(\frac{1}{\cos x} - \cos x\right).
$$

This gives

$$
\sin^2 x = 1 - \cos^2 x.
$$

**Step 5.** Recognize that the Pythagorean identity tells us

$$
\sin^2 x + \cos^2 x = 1,
$$

or equivalently,

$$
\sin^2 x = 1 - \cos^2 x.
$$

Thus, the equation reduces to an identity.

**Step 6.** Take into account the domain restrictions. In our steps, we required:
- $$\tan x$$ to be defined, which means $$\cos x \neq 0$$.
- We divided by $$\sin x$$, so $$\sin x \neq 0$$.

So the solution is any $$x$$ for which
$$
\cos x \neq 0 \quad \text{and} \quad \sin x \neq 0.
$$

This means

$$
x \neq \frac{\pi}{2} + n\pi \quad \text{and} \quad x \neq n\pi \quad \text{for any integer } n.
$$

**Final Answer:** The equation is an identity. Its solutions are all $$x$$ satisfying

$$
x \in \mathbb{R} \setminus \{ n\pi,\, \frac{\pi}{2}+n\pi \mid n\in\mathbb{Z}\}.
$$
The equation is an identity, meaning it holds true for all real numbers $$ x $$ except where $$ \sin x = 0 $$ or $$ \cos x = 0 $$.

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