The number of solution of the equation `sqrt(13-18tanx)=6 tan x-3`, where `-2pilt x lt 2pi`is

Options

A. 0
B. 2
C. 4
D. 8

Explanation

To solve the equation \(\sqrt{13 - 18 \tan x} = 6 \tan x - 3\) for the number of solutions in the interval \(-2\pi < x < 2\pi\), we will follow these steps:

### Step 1: Square both sides of the equation
Start by squaring both sides to eliminate the square root:
\[ 13 - 18 \tan x = (6 \tan x - 3)^2 \]

### Step 2: Expand the right side
Using the identity \((a - b)^2 = a^2 - 2ab + b^2\):
\[ (6 \tan x - 3)^2 = 36 \tan^2 x - 36 \tan x + 9 \]
So, the equation becomes:
\[ 13 - 18 \tan x = 36 \tan^2 x - 36 \tan x + 9 \]

### Step 3: Rearrange the equation
Rearranging gives:
\[ 0 = 36 \tan^2 x - 36 \tan x + 9 + 18 \tan x - 13 \]
This simplifies to:
\[ 0 = 36 \tan^2 x - 18 \tan x - 4 \]

### Step 4: Divide the entire equation by 2
To simplify:
\[ 0 = 18 \tan^2 x - 9 \tan x - 2 \]

### Step 5: Use the quadratic formula
Let \(y = \tan x\). The equation now is:
\[ 18y^2 - 9y - 2 = 0 \]
Using the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ y = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 18 \cdot (-2)}}{2 \cdot 18} \]
Calculating the discriminant:
\[ b^2 - 4ac = 81 + 144 = 225 \]
Thus, we have:
\[ y = \frac{9 \pm 15}{36} \]
Calculating the two possible values:
1. \(y_1 = \frac{24}{36} = \frac{2}{3}\)
2. \(y_2 = \frac{-6}{36} = -\frac{1}{6}\)

### Step 6: Find the angles for \(\tan x\)
Now we find the values of \(x\) for each \(y\):
1. For \(y_1 = \frac{2}{3}\):
\[ x = \tan^{-1}\left(\frac{2}{3}\right) + n\pi \]
2. For \(y_2 = -\frac{1}{6}\):
\[ x = \tan^{-1}\left(-\frac{1}{6}\right) + n\pi \]

### Step 7: Determine the number of solutions in the interval \(-2\pi < x < 2\pi\)
For \(y_1 = \frac{2}{3}\):
- \(n = 0\): \(x_1 = \tan^{-1}\left(\frac{2}{3}\right)\)
- \(n = 1\): \(x_2 = \tan^{-1}\left(\frac{2}{3}\right) + \pi\)
- \(n = -1\): \(x_3 = \tan^{-1}\left(\frac{2}{3}\right) - \pi\)
- \(n = -2\): \(x_4 = \tan^{-1}\left(\frac{2}{3}\right) - 2\pi\)

For \(y_2 = -\frac{1}{6}\):
- \(n = 0\): \(x_5 = \tan^{-1}\left(-\frac{1}{6}\right)\)
- \(n = 1\): \(x_6 = \tan^{-1}\left(-\frac{1}{6}\right) + \pi\)
- \(n = -1\): \(x_7 = \tan^{-1}\left(-\frac{1}{6}\right) - \pi\)
- \(n = -2\): \(x_8 = \tan^{-1}\left(-\frac{1}{6}\right) - 2\pi\)

### Step 8: Count the valid solutions
Each value of \(y\) gives us 4 solutions in the interval \(-2\pi < x < 2\pi\). Thus, we have:
- From \(y_1\): 4 solutions
- From \(y_2\): 4 solutions

### Conclusion
The total number of solutions is \(4 + 4 = 8\).

### Final Answer
The number of solutions of the equation \(\sqrt{13 - 18 \tan x} = 6 \tan x - 3\) in the interval \(-2\pi < x < 2\pi\) is **8**.