Let `(b cos x)/(2 cos 2x-1)=(b + sin x)/((cos^(2) x-3 sin^(2) x) tan x), b in R`.
Equation has solutions if

Answer

To solve the equation

\[ \frac{b \cos x}{2 \cos 2x - 1} = \frac{b + \sin x}{(\cos^2 x - 3 \sin^2 x) \tan x} \]

where \( b \in \mathbb{R} \), we need to analyze the conditions under which this equation has solutions.

### Step 1: Identify Denominators
First, we need to ensure that the denominators are not equal to zero.

1. **Condition from \( 2 \cos 2x - 1 \neq 0 \)**:
\[ 2 \cos 2x - 1 \neq 0 \implies \cos 2x \neq \frac{1}{2} \]
This occurs when:
\[ 2x \neq n\pi \pm \frac{\pi}{3} \implies x \neq \frac{n\pi}{2} \pm \frac{\pi}{6} \]
where \( n \in \mathbb{Z} \).

2. **Condition from \( \cos^2 x - 3 \sin^2 x \neq 0 \)**:
\[ \cos^2 x - 3 \sin^2 x \neq 0 \implies \cos^2 x \neq 3 \sin^2 x \]
This can be rewritten using \( \sin^2 x = 1 - \cos^2 x \):
\[ \cos^2 x \neq 3(1 - \cos^2 x) \implies 4\cos^2 x \neq 3 \implies \cos^2 x \neq \frac{3}{4} \]
Thus, \( \cos x \neq \pm \frac{\sqrt{3}}{2} \).

3. **Condition from \( \tan x \neq 0 \)**:
\[ \tan x \neq 0 \implies x \neq n\pi \]
where \( n \in \mathbb{Z} \).

### Step 2: Rewrite the Equation
Now, we can rewrite the original equation:
\[ b \cos x ( \cos^2 x - 3 \sin^2 x) \tan x = (b + \sin x)(2 \cos 2x - 1) \]

### Step 3: Simplify the Equation
Using the identity \( \cos 2x = 2\cos^2 x - 1 \):
\[ b \cos x ( \cos^2 x - 3 \sin^2 x) \frac{\sin x}{\cos x} = (b + \sin x)(2\cos^2 x - 1 - 1) \]
This simplifies to:
\[ b \sin x (\cos^2 x - 3 \sin^2 x) = (b + \sin x)(2\cos^2 x - 2) \]

### Step 4: Analyze the Resulting Equation
We can cancel \( \cos^2 x - 3 \sin^2 x \) from both sides (assuming it is not zero):
\[ b \sin x = (b + \sin x)(2) \]
This leads to:
\[ b \sin x = 2b + 2\sin x \]
Rearranging gives:
\[ (b - 2)\sin x = 2b \]

### Step 5: Solve for \( \sin x \)
From the equation:
\[ \sin x = \frac{2b}{b - 2} \]
For this to have solutions, \( \sin x \) must satisfy:
\[ -1 \leq \frac{2b}{b - 2} \leq 1 \]

### Step 6: Solve the Inequalities
1. **First Inequality**:
\[ 2b \geq - (b - 2) \implies 3b \geq 2 \implies b \geq \frac{2}{3} \]

2. **Second Inequality**:
\[ 2b \leq b - 2 \implies b \leq -2 \]

### Step 7: Combine the Results
Thus, the values of \( b \) for which the equation has solutions are:
\[ b \in (-\infty, -2] \cup \left[\frac{2}{3}, \infty\right) \]

### Final Answer
The equation has solutions if:
\[ b \in (-\infty, -2] \cup \left[\frac{2}{3}, \infty\right) \]