In reducing a given trignometric equation to the standard form (sinx `=sin alpha)` or `( cosx = cos alpha)` etc. We apply several trignometric or Algebric transformation.As a result of which final form so obtained may not be equivalent to the original equation resulting either in loss of solutions or appearance of extraneous solutions.
The solution of `sqrt(13-18tanx) = 6 tanx -3` is

Options

A. (A)`tan x=-1/6`
B. (B)`tan x=2/3`
C. (C)`tan x (-infty, -1/6) cup (2/3, infty)`
D. (D)`tan x=3/4`

Answer

To solve the equation \( \sqrt{13 - 18 \tan x} = 6 \tan x - 3 \), we will follow these steps:

### Step 1: Square both sides
To eliminate the square root, we square both sides of the equation:

\[ (\sqrt{13 - 18 \tan x})^2 = (6 \tan x - 3)^2 \]

This simplifies to:

\[ 13 - 18 \tan x = (6 \tan x - 3)(6 \tan x - 3) \]

### Step 2: Expand the right side
Now, we expand the right side:

\[ (6 \tan x - 3)^2 = 36 \tan^2 x - 36 \tan x + 9 \]

So, we have:

\[ 13 - 18 \tan x = 36 \tan^2 x - 36 \tan x + 9 \]

### Step 3: Rearrange the equation
Next, we rearrange the equation to bring all terms to one side:

\[ 36 \tan^2 x - 36 \tan x + 9 + 18 \tan x - 13 = 0 \]

This simplifies to:

\[ 36 \tan^2 x - 18 \tan x - 4 = 0 \]

### Step 4: Factor the quadratic equation
Now, we can factor the quadratic equation. We can rewrite it as:

\[ 36 \tan^2 x - 24 \tan x + 6 \tan x - 4 = 0 \]

Factoring by grouping gives us:

\[ 12 \tan x (3 \tan x - 2) + 2(3 \tan x - 2) = 0 \]

This can be factored as:

\[ (12 \tan x + 2)(3 \tan x - 2) = 0 \]

### Step 5: Solve for \( \tan x \)
Setting each factor to zero gives us:

1. \( 12 \tan x + 2 = 0 \)
\[ \tan x = -\frac{1}{6} \]

2. \( 3 \tan x - 2 = 0 \)
\[ \tan x = \frac{2}{3} \]

### Step 6: General solutions
The general solutions for \( \tan x = -\frac{1}{6} \) and \( \tan x = \frac{2}{3} \) can be expressed as:

1. For \( \tan x = -\frac{1}{6} \):
\[ x = n\pi + \tan^{-1}\left(-\frac{1}{6}\right), \quad n \in \mathbb{Z} \]

2. For \( \tan x = \frac{2}{3} \):
\[ x = n\pi + \tan^{-1}\left(\frac{2}{3}\right), \quad n \in \mathbb{Z} \]

### Final Answer
Thus, the solutions to the equation \( \sqrt{13 - 18 \tan x} = 6 \tan x - 3 \) are:

\[ x = n\pi + \tan^{-1}\left(-\frac{1}{6}\right) \quad \text{and} \quad x = n\pi + \tan^{-1}\left(\frac{2}{3}\right), \quad n \in \mathbb{Z} \]