The equation `sin^(6)x+cos^(6)x=a`, has a real solution if

Options

A. `1//2leale1`
B. `1//4leale1`
C. `-1leale1`
D. `0leale1//2`

Answer

To solve the equation \( \sin^6 x + \cos^6 x = a \) and determine the conditions under which it has real solutions, we can follow these steps:

### Step 1: Rewrite the expression
We start by rewriting \( \sin^6 x + \cos^6 x \) using the identity for the sum of cubes:
\[ \sin^6 x + \cos^6 x = (\sin^2 x)^3 + (\cos^2 x)^3 \]
Using the formula \( A^3 + B^3 = (A + B)(A^2 - AB + B^2) \), where \( A = \sin^2 x \) and \( B = \cos^2 x \), we have:
\[ \sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)((\sin^2 x)^2 - \sin^2 x \cos^2 x + (\cos^2 x)^2) \]

### Step 2: Simplify using the Pythagorean identity
Since \( \sin^2 x + \cos^2 x = 1 \), we can simplify further:
\[ \sin^6 x + \cos^6 x = 1 \cdot ((\sin^2 x)^2 + (\cos^2 x)^2 - \sin^2 x \cos^2 x) \]

### Step 3: Use the identity for squares
Now, we can express \( (\sin^2 x)^2 + (\cos^2 x)^2 \) using the identity:
\[ (\sin^2 x + \cos^2 x)^2 = \sin^4 x + \cos^4 x + 2\sin^2 x \cos^2 x \]
Thus,
\[ \sin^4 x + \cos^4 x = 1 - 2\sin^2 x \cos^2 x \]
Substituting this back, we get:
\[ \sin^6 x + \cos^6 x = 1 - 3\sin^2 x \cos^2 x \]

### Step 4: Express in terms of \( \sin 2x \)
Using the identity \( \sin 2x = 2\sin x \cos x \), we can express \( \sin^2 x \cos^2 x \) as:
\[ \sin^2 x \cos^2 x = \frac{\sin^2 2x}{4} \]
Substituting this into our expression gives:
\[ \sin^6 x + \cos^6 x = 1 - \frac{3}{4} \sin^2 2x \]

### Step 5: Determine the range of \( a \)
Now we have:
\[ \sin^6 x + \cos^6 x = 1 - \frac{3}{4} \sin^2 2x \]
The maximum value of \( \sin^2 2x \) is 1, thus:
\[ 1 - \frac{3}{4} \cdot 1 = \frac{1}{4} \]
The minimum value occurs when \( \sin^2 2x = 0 \):
\[ 1 - \frac{3}{4} \cdot 0 = 1 \]
Thus, we find that:
\[ \frac{1}{4} \leq a \leq 1 \]

### Conclusion
The equation \( \sin^6 x + \cos^6 x = a \) has real solutions if:
\[ \frac{1}{4} \leq a \leq 1 \]