The set of values of a for which the equation `Sin^4 x + Cos^4 x = a` has a solution is

Answer

To solve the equation \( \sin^4 x + \cos^4 x = a \) and find the set of values of \( a \) for which this equation has a solution, we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ \sin^4 x + \cos^4 x \]
We can use the identity for the sum of squares:
\[ a^2 + b^2 = (a + b)^2 - 2ab \]
Letting \( a = \sin^2 x \) and \( b = \cos^2 x \), we have:
\[ \sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x \]
Since \( \sin^2 x + \cos^2 x = 1 \), we can simplify this to:
\[ \sin^4 x + \cos^4 x = 1 - 2\sin^2 x \cos^2 x \]

### Step 2: Use the double angle identity
We know that:
\[ \sin^2 x \cos^2 x = \frac{1}{4} \sin^2(2x) \]
Thus, we can rewrite our expression as:
\[ \sin^4 x + \cos^4 x = 1 - \frac{1}{2} \sin^2(2x) \]

### Step 3: Determine the range of the expression
The function \( \sin^2(2x) \) varies from 0 to 1. Therefore:
\[ 0 \leq \sin^2(2x) \leq 1 \]
This implies:
\[ 0 \leq \frac{1}{2} \sin^2(2x) \leq \frac{1}{2} \]
Consequently, we have:
\[ 1 - \frac{1}{2} \leq \sin^4 x + \cos^4 x \leq 1 - 0 \]
This simplifies to:
\[ \frac{1}{2} \leq \sin^4 x + \cos^4 x \leq 1 \]

### Step 4: Set the range for \( a \)
Thus, the set of values for \( a \) for which the equation \( \sin^4 x + \cos^4 x = a \) has a solution is:
\[ \frac{1}{2} \leq a \leq 1 \]

### Final Answer
The set of values of \( a \) is:
\[ a \in \left[\frac{1}{2}, 1\right] \]

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