Solve `7 cos^(2)x+sin x cos x-3=0`.

Explanation

To solve the equation \( 7 \cos^2 x + \sin x \cos x - 3 = 0 \), we can follow these steps:

### Step 1: Rewrite the equation in terms of sine and cosine
We start with the equation:
\[ 7 \cos^2 x + \sin x \cos x - 3 = 0 \]

### Step 2: Divide the entire equation by \( \cos^2 x \)
Dividing by \( \cos^2 x \) (assuming \( \cos x \neq 0 \)):
\[ 7 + \tan x - 3 \tan^2 x = 0 \]

### Step 3: Rearrange the equation
Rearranging gives us:
\[ 3 \tan^2 x - \tan x - 4 = 0 \]

### Step 4: Factor the quadratic equation
We can factor the quadratic equation:
\[ (3 \tan x + 4)(\tan x - 1) = 0 \]

### Step 5: Solve for \( \tan x \)
Setting each factor to zero gives us:
1. \( 3 \tan x + 4 = 0 \) which leads to \( \tan x = -\frac{4}{3} \)
2. \( \tan x - 1 = 0 \) which leads to \( \tan x = 1 \)

### Step 6: Find the general solutions for \( x \)
For \( \tan x = -\frac{4}{3} \):
\[ x = \tan^{-1}\left(-\frac{4}{3}\right) + n\pi \quad (n \in \mathbb{Z}) \]

For \( \tan x = 1 \):
\[ x = \frac{\pi}{4} + n\pi \quad (n \in \mathbb{Z}) \]

### Final Solutions
Thus, the complete solution set is:
1. \( x = \tan^{-1}\left(-\frac{4}{3}\right) + n\pi \)
2. \( x = \frac{\pi}{4} + n\pi \)

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