If the expression `([s in(x/2)+cos(x/2)-i t a n(x)])/([1+2is in(x/2)])` is real, then the set of all possible values of `x` is.........

Options

A. `2n pi + 2 "tan"^(-1)k, k in R, n in Z`
B. `2n pi + 2 "tan"^(-1)k," where " k in(0, 1), n in Z`
C. `2n pi + 2"tan"^(-1)k, "where" k in (1, 2), n in Z`
D. `2n pi + 2"tan"^(-1) k, k in (2, 3), n in Z`

Answer

To solve the problem, we need to determine the values of \( x \) for which the expression

\[ \frac{\sin\left(\frac{x}{2}\right) + \cos\left(\frac{x}{2}\right) - i \tan(x)}{1 + 2i \sin\left(\frac{x}{2}\right)} \]

is real.

### Step 1: Set the expression to be real
For the expression to be real, the imaginary part must equal zero.

### Step 2: Rationalize the denominator
We can multiply the numerator and denominator by the conjugate of the denominator:

\[ \frac{(\sin\left(\frac{x}{2}\right) + \cos\left(\frac{x}{2}\right) - i \tan(x))(1 - 2i \sin\left(\frac{x}{2}\right))}{(1 + 2i \sin\left(\frac{x}{2}\right))(1 - 2i \sin\left(\frac{x}{2}\right))} \]

The denominator simplifies to:

\[ 1 + 4 \sin^2\left(\frac{x}{2}\right) \]

### Step 3: Expand the numerator
Now, we expand the numerator:

\[ (\sin\left(\frac{x}{2}\right) + \cos\left(\frac{x}{2}\right))(1 - 2i \sin\left(\frac{x}{2}\right)) - i \tan(x)(1 - 2i \sin\left(\frac{x}{2}\right)) \]

This gives us:

\[ \sin\left(\frac{x}{2}\right) + \cos\left(\frac{x}{2}\right) - 2i \sin^2\left(\frac{x}{2}\right) - i \tan(x) + 2 \tan(x) \sin\left(\frac{x}{2}\right) \]

### Step 4: Collect real and imaginary parts
The real part is:

\[ \sin\left(\frac{x}{2}\right) + \cos\left(\frac{x}{2}\right) \]

The imaginary part is:

\[ -2 \sin^2\left(\frac{x}{2}\right) - \tan(x) + 2 \tan(x) \sin\left(\frac{x}{2}\right) \]

### Step 5: Set the imaginary part to zero
To find when the expression is real, we set the imaginary part to zero:

\[ -2 \sin^2\left(\frac{x}{2}\right) - \tan(x) + 2 \tan(x) \sin\left(\frac{x}{2}\right) = 0 \]

### Step 6: Substitute \(\tan(x)\)
Using the identity \(\tan(x) = \frac{\sin(x)}{\cos(x)}\), we can rewrite the equation as:

\[ -2 \sin^2\left(\frac{x}{2}\right) - \frac{\sin(x)}{\cos(x)} + 2 \frac{\sin(x)}{\cos(x)} \sin\left(\frac{x}{2}\right) = 0 \]

### Step 7: Simplify the equation
Using the double angle identity:

\[ \sin(x) = 2 \sin\left(\frac{x}{2}\right) \cos\left(\frac{x}{2}\right) \]

We can substitute this into our equation and simplify further.

### Step 8: Solve the resulting equation
After simplification, we will arrive at a polynomial equation in terms of \(\tan\left(\frac{x}{2}\right)\).

### Step 9: Find the roots
By solving this polynomial equation, we can find the values of \(\tan\left(\frac{x}{2}\right)\) and subsequently find \(x\).

### Step 10: General solution
The general solution will be of the form:

\[ x = 2n\pi + 2\tan^{-1}(k) \]

where \(k\) are the roots found in the previous step.

### Final Answer
The set of all possible values of \(x\) is:

\[ x = 2n\pi + 2\tan^{-1}(k) \quad \text{for } k \text{ in the range of roots found} \]