If `((1-asinx)/(1+asinx))sqrt((1+2asinx)/(1-2asinx))=1` , is where `a in R` then (a) `x in varphi` (b) `x in R\ AA\ a` (c) `a=0,\ x in R` (d) `a\ in R ,\ x\ in npi,` where `n in I`

Answer

To solve the equation

\[ \frac{1 - a \sin x}{1 + a \sin x} \sqrt{\frac{1 + 2a \sin x}{1 - 2a \sin x}} = 1, \]

where \( a \in \mathbb{R} \), we will follow these steps:

### Step 1: Rearranging the Equation
We start by isolating the square root:

\[ \frac{1 - a \sin x}{1 + a \sin x} = \sqrt{\frac{1 + 2a \sin x}{1 - 2a \sin x}}. \]

### Step 2: Squaring Both Sides
Next, we square both sides to eliminate the square root:

\[ \left(\frac{1 - a \sin x}{1 + a \sin x}\right)^2 = \frac{1 + 2a \sin x}{1 - 2a \sin x}. \]

### Step 3: Expanding Both Sides
Expanding the left side using the formula \((\frac{a}{b})^2 = \frac{a^2}{b^2}\):

\[ \frac{(1 - a \sin x)^2}{(1 + a \sin x)^2} = \frac{1 + 2a \sin x}{1 - 2a \sin x}. \]

### Step 4: Cross Multiplying
Cross-multiplying gives us:

\[ (1 - a \sin x)^2 (1 - 2a \sin x) = (1 + 2a \sin x)(1 + a \sin x)^2. \]

### Step 5: Expanding Both Sides
Now we expand both sides:

Left Side:
\[ (1 - 2a \sin x + a^2 \sin^2 x)(1 - 2a \sin x) = 1 - 2a \sin x + a^2 \sin^2 x - 2a \sin x + 4a^2 \sin^2 x - 2a^3 \sin^3 x. \]

Right Side:
\[ (1 + 2a \sin x)(1 + 2a \sin x + a^2 \sin^2 x) = 1 + 2a \sin x + a^2 \sin^2 x + 2a \sin x + 4a^2 \sin^2 x + 2a^3 \sin^3 x. \]

### Step 6: Simplifying Both Sides
After simplification, we equate both sides. This leads to a polynomial equation in terms of \( \sin x \).

### Step 7: Solving for \( \sin x \)
We can factor the equation to find the values of \( \sin x \):

\[ a^2 \sin^2 x + 4a^2 \sin^2 x - 2a^3 \sin^3 x = 0. \]

Factoring out \( \sin x \):

\[ \sin x (a^2 + 4a^2 \sin x - 2a^3 \sin^2 x) = 0. \]

### Step 8: Finding Solutions
This gives us two cases:
1. \( \sin x = 0 \)
2. The quadratic equation \( a^2 + 4a^2 \sin x - 2a^3 \sin^2 x = 0 \).

### Step 9: Analyzing the Cases
- From \( \sin x = 0 \), we find \( x = n\pi \) where \( n \in \mathbb{Z} \).
- For \( a = 0 \), any \( x \) is valid.

### Conclusion
The valid options are:
- (c) \( a = 0, x \in \mathbb{R} \)
- (d) \( a \in \mathbb{R}, x \in n\pi \)

Thus, the correct answer is option (d).