Show that the equation `sintheta=x+1/x` is not possible if `x` is real.

Answer

To show that the equation \( \sin \theta = x + \frac{1}{x} \) is not possible if \( x \) is real, we can follow these steps:

### Step 1: Start with the given equation
We have the equation:
\[ \sin \theta = x + \frac{1}{x} \]

### Step 2: Square both sides
To eliminate the fraction, we square both sides:
\[ \sin^2 \theta = \left( x + \frac{1}{x} \right)^2 \]

### Step 3: Expand the right-hand side
Using the identity \( (a + b)^2 = a^2 + 2ab + b^2 \), we expand the right-hand side:
\[ \sin^2 \theta = x^2 + 2 + \frac{1}{x^2} \]

### Step 4: Rearrange the equation
Rearranging gives us:
\[ \sin^2 \theta = x^2 + 2 + \frac{1}{x^2} \]

### Step 5: Express \( x^2 + \frac{1}{x^2} \)
We can express \( x^2 + \frac{1}{x^2} \) in terms of \( x \):
\[ x^2 + \frac{1}{x^2} = \left( x - \frac{1}{x} \right)^2 + 2 \]
Thus, we can rewrite our equation:
\[ \sin^2 \theta = \left( x - \frac{1}{x} \right)^2 + 4 \]

### Step 6: Analyze the right-hand side
The term \( \left( x - \frac{1}{x} \right)^2 \) is always non-negative (since it is a square). Therefore, the smallest value of \( \sin^2 \theta \) can take is:
\[ \sin^2 \theta \geq 4 \]

### Step 7: Compare with the range of \( \sin^2 \theta \)
We know that:
\[ \sin^2 \theta \leq 1 \]
for any real value of \( \theta \).

### Step 8: Conclusion
Since \( \sin^2 \theta \) cannot be greater than 1, but we derived that \( \sin^2 \theta \geq 4 \), we conclude that:
\[ \sin \theta = x + \frac{1}{x} \text{ is not possible for real } x. \]