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 set of the equation `sqrt(5-2sinx) = 6 sinx -1` is

Options

A. `n pi+(-1)^(n)` where `sin theta = -2/9` or `x= npi +(-1)^(n) pi/6 n in I`
B. `x=npi + (-1)^(n) pi/6, n in I`
C. `x=npi +(-1)^(n) theta`, where `sin theta=2/9`
D. Null set

Answer

To solve the equation \( \sqrt{5 - 2\sin x} = 6\sin x - 1 \), we will follow these steps:

### Step 1: Square both sides
To eliminate the square root, we square both sides of the equation:
\[ (\sqrt{5 - 2\sin x})^2 = (6\sin x - 1)^2 \]
This simplifies to:
\[ 5 - 2\sin x = (6\sin x - 1)(6\sin x - 1) \]

### Step 2: Expand the right-hand side
Using the formula \( (a - b)^2 = a^2 - 2ab + b^2 \), we expand the right-hand side:
\[ 5 - 2\sin x = 36\sin^2 x - 12\sin x + 1 \]

### Step 3: Rearrange the equation
Now, we rearrange the equation to bring all terms to one side:
\[ 0 = 36\sin^2 x - 12\sin x + 1 + 2\sin x - 5 \]
This simplifies to:
\[ 0 = 36\sin^2 x - 10\sin x - 4 \]

### Step 4: Factor or use the quadratic formula
We can now use the quadratic formula \( \sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 36, b = -10, c = -4 \).

Calculating the discriminant \( D \):
\[ D = (-10)^2 - 4 \cdot 36 \cdot (-4) = 100 + 576 = 676 \]
Now, applying the quadratic formula:
\[ \sin x = \frac{-(-10) \pm \sqrt{676}}{2 \cdot 36} = \frac{10 \pm 26}{72} \]

### Step 5: Calculate the values of \( \sin x \)
Calculating the two possible values:
1. \( \sin x = \frac{36}{72} = \frac{1}{2} \)
2. \( \sin x = \frac{-16}{72} = -\frac{2}{9} \)

### Step 6: Check for extraneous solutions
Since the original equation involves a square root, we must check if these solutions are valid:
- For \( \sin x = \frac{1}{2} \):
\[ 6\sin x - 1 = 6 \cdot \frac{1}{2} - 1 = 3 - 1 = 2 \quad \text{(valid)} \]
- For \( \sin x = -\frac{2}{9} \):
\[ 6\sin x - 1 = 6 \cdot -\frac{2}{9} - 1 = -\frac{12}{9} - 1 = -\frac{12}{9} - \frac{9}{9} = -\frac{21}{9} \quad \text{(not valid)} \]

Thus, the only valid solution is \( \sin x = \frac{1}{2} \).

### Step 7: Find the general solution
The general solution for \( \sin x = \frac{1}{2} \) is:
\[ x = n\pi + (-1)^n \frac{\pi}{6}, \quad n \in \mathbb{Z} \]

### Final Answer
The solution set of the equation \( \sqrt{5 - 2\sin x} = 6\sin x - 1 \) is:
\[ x = n\pi + (-1)^n \frac{\pi}{6}, \quad n \in \mathbb{Z} \]