The values of `alpha` for which the equation `alpha^2/(1-tan^2x)=(sin^2+xalpha^2-2)/(cos2x)` has solution can be

Options

A. `alpha le -1 `
B. `alpha ge 1`
C. `alpha =1//2`
D. `alpha ` in any real number

Answer

To solve the equation

\[ \frac{\alpha^2}{1 - \tan^2 x} = \frac{\sin^2 x + \alpha^2 - 2}{\cos 2x}, \]

we will follow these steps:

### Step 1: Rewrite the equation

Start by rewriting the equation in a more manageable form. Recall that

\[ \tan^2 x = \frac{\sin^2 x}{\cos^2 x} \quad \text{and} \quad \cos 2x = \cos^2 x - \sin^2 x. \]

This allows us to express the left-hand side as:

\[ \frac{\alpha^2 \cos^2 x}{\cos^2 x - \sin^2 x}. \]

Thus, the equation becomes:

\[ \frac{\alpha^2 \cos^2 x}{\cos^2 x - \sin^2 x} = \frac{\sin^2 x + \alpha^2 - 2}{\cos^2 x - \sin^2 x}. \]

### Step 2: Clear the denominators

Since both sides of the equation have the same denominator, we can multiply both sides by \((\cos^2 x - \sin^2 x)\) (assuming it is not zero):

\[ \alpha^2 \cos^2 x = \sin^2 x + \alpha^2 - 2. \]

### Step 3: Rearrange the equation

Rearranging gives:

\[ \alpha^2 \cos^2 x - \sin^2 x - \alpha^2 + 2 = 0. \]

### Step 4: Factor the equation

This can be rearranged as:

\[ \alpha^2 (\cos^2 x - 1) + 2 - \sin^2 x = 0. \]

Using the identity \(\cos^2 x = 1 - \sin^2 x\), we can substitute:

\[ \alpha^2 (-\sin^2 x) + 2 - \sin^2 x = 0. \]

### Step 5: Combine like terms

Combining the terms gives:

\[ (-\alpha^2 - 1) \sin^2 x + 2 = 0. \]

### Step 6: Solve for \(\sin^2 x\)

Rearranging this gives:

\[ \sin^2 x = \frac{2}{\alpha^2 + 1}. \]

### Step 7: Determine the range of \(\sin^2 x\)

Since \(\sin^2 x\) must lie in the interval \([0, 1]\), we need:

\[ 0 \leq \frac{2}{\alpha^2 + 1} \leq 1. \]

### Step 8: Solve the inequalities

1. From \(0 \leq \frac{2}{\alpha^2 + 1}\), we see that \(\alpha^2 + 1 > 0\) is always true.
2. From \(\frac{2}{\alpha^2 + 1} \leq 1\):

\[ 2 \leq \alpha^2 + 1 \implies \alpha^2 \geq 1. \]

### Step 9: Find the values of \(\alpha\)

This means:

\[ \alpha \leq -1 \quad \text{or} \quad \alpha \geq 1. \]

### Conclusion

Thus, the values of \(\alpha\) for which the equation has solutions are:

\[ \alpha \leq -1 \quad \text{or} \quad \alpha \geq 1. \]