Solve `( sin x + "i" cos x )/( 1 +i),i =sqrt(-1)` when it is purely imaginary .

Answer

To solve the equation \(\frac{\sin x + i \cos x}{1 + i}\) when it is purely imaginary, we will follow these steps:

### Step 1: Rationalize the denominator
We start with the expression:
\[ \frac{\sin x + i \cos x}{1 + i} \]
To eliminate the imaginary unit in the denominator, we multiply the numerator and denominator by the conjugate of the denominator, which is \(1 - i\):
\[ \frac{(\sin x + i \cos x)(1 - i)}{(1 + i)(1 - i)} \]

### Step 2: Simplify the denominator
Calculating the denominator:
\[ (1 + i)(1 - i) = 1^2 - i^2 = 1 - (-1) = 1 + 1 = 2 \]

### Step 3: Expand the numerator
Now we expand the numerator:
\[ (\sin x + i \cos x)(1 - i) = \sin x(1 - i) + i \cos x(1 - i) = \sin x - i \sin x + i \cos x - \cos x \]
Combining the terms:
\[ = (\sin x - \cos x) + i(\cos x - \sin x) \]

### Step 4: Write the full expression
Now we can write the full expression:
\[ \frac{(\sin x - \cos x) + i(\cos x - \sin x)}{2} \]
This can be separated into real and imaginary parts:
\[ = \frac{\sin x - \cos x}{2} + i \frac{\cos x - \sin x}{2} \]

### Step 5: Set the real part to zero
Since we want the expression to be purely imaginary, we set the real part equal to zero:
\[ \frac{\sin x - \cos x}{2} = 0 \]
This simplifies to:
\[ \sin x - \cos x = 0 \]

### Step 6: Solve for \(x\)
This implies:
\[ \sin x = \cos x \]
Dividing both sides by \(\cos x\) (assuming \(\cos x \neq 0\)):
\[ \tan x = 1 \]
The general solution for this equation is:
\[ x = n\pi + \frac{\pi}{4}, \quad n \in \mathbb{Z} \]

### Step 7: Final answer
Thus, the solution is:
\[ x = n\pi + \frac{\pi}{4}, \quad n \in \mathbb{Z} \]