The equation `"sin"^(4) theta + "cos"^(4) theta = a` has a real solution if

Options

A. `a in [1//2, 1]`
B. ` a in [1//4, 1//2]`
C. `a in [1//3, 1]`
D. none of these

Answer

To solve the equation \( \sin^4 \theta + \cos^4 \theta = a \) and find the conditions under which it has a real solution, we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \sin^4 \theta + \cos^4 \theta = a \]
We can rewrite \( \sin^4 \theta + \cos^4 \theta \) using the identity:
\[ \sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta \]
Since \( \sin^2 \theta + \cos^2 \theta = 1 \), we have:
\[ \sin^4 \theta + \cos^4 \theta = 1 - 2\sin^2 \theta \cos^2 \theta \]

### Step 2: Substitute for \( \sin^2 \theta \cos^2 \theta \)
Recall that \( \sin^2 \theta \cos^2 \theta = \frac{1}{4} \sin^2(2\theta) \). Thus, we can substitute this into our equation:
\[ \sin^4 \theta + \cos^4 \theta = 1 - 2 \cdot \frac{1}{4} \sin^2(2\theta) = 1 - \frac{1}{2} \sin^2(2\theta) \]

### Step 3: Set the equation equal to \( a \)
Now we have:
\[ 1 - \frac{1}{2} \sin^2(2\theta) = a \]
Rearranging gives:
\[ \frac{1}{2} \sin^2(2\theta) = 1 - a \]
Thus,
\[ \sin^2(2\theta) = 2(1 - a) \]

### Step 4: Determine the range of \( a \)
The sine function has a range of values between 0 and 1. Therefore, we need:
\[ 0 \leq \sin^2(2\theta) \leq 1 \]
This leads to:
\[ 0 \leq 2(1 - a) \leq 1 \]

### Step 5: Solve the inequalities
From \( 0 \leq 2(1 - a) \), we get:
\[ 1 - a \geq 0 \quad \Rightarrow \quad a \leq 1 \]
From \( 2(1 - a) \leq 1 \), we get:
\[ 1 - a \leq \frac{1}{2} \quad \Rightarrow \quad a \geq \frac{1}{2} \]

### Conclusion
Combining these results, we find that \( a \) must satisfy:
\[ \frac{1}{2} \leq a \leq 1 \]
Thus, the equation \( \sin^4 \theta + \cos^4 \theta = a \) has a real solution if:
\[ a \in \left[\frac{1}{2}, 1\right] \]