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

Options

A. `lambda in [1//2, 1]`
B. `lambda in [1//4, 1]`
C. `lambda in [-1, 1]`
D. `lambda in [0, 1//2]`

Answer

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

### Step 1: Rewrite the Equation
We start with the equation:
\[ \sin^6 x + \cos^6 x = \lambda \]
This can be rewritten using the identity for cubes:
\[ \sin^6 x + \cos^6 x = (\sin^2 x)^3 + (\cos^2 x)^3 \]

### Step 2: Apply the Sum of Cubes Formula
Using the formula for the sum of cubes, \( a^3 + b^3 = (a + b)(a^2 + b^2 - ab) \), we set \( a = \sin^2 x \) and \( b = \cos^2 x \):
\[ \sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)((\sin^2 x)^2 + (\cos^2 x)^2 - \sin^2 x \cos^2 x) \]

### Step 3: Simplify Using Trigonometric Identity
Since \( \sin^2 x + \cos^2 x = 1 \):
\[ \sin^6 x + \cos^6 x = 1 \cdot ((\sin^2 x)^2 + (\cos^2 x)^2 - \sin^2 x \cos^2 x) \]
Now, we need to simplify \( (\sin^2 x)^2 + (\cos^2 x)^2 \):
\[ (\sin^2 x)^2 + (\cos^2 x)^2 = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x \]
Thus,
\[ \sin^6 x + \cos^6 x = 1 - 3\sin^2 x \cos^2 x \]

### Step 4: Substitute Back into the Equation
Now we can substitute this back into our equation:
\[ 1 - 3\sin^2 x \cos^2 x = \lambda \]

### Step 5: Express \( \sin^2 x \cos^2 x \) in Terms of \( \lambda \)
Rearranging gives:
\[ 3\sin^2 x \cos^2 x = 1 - \lambda \]
This implies:
\[ \sin^2 x \cos^2 x = \frac{1 - \lambda}{3} \]

### Step 6: Use the Identity for \( \sin^2 2x \)
We know that \( \sin^2 x \cos^2 x = \frac{1}{4} \sin^2 2x \):
\[ \frac{1}{4} \sin^2 2x = \frac{1 - \lambda}{3} \]
Multiplying both sides by 4:
\[ \sin^2 2x = \frac{4(1 - \lambda)}{3} \]

### Step 7: Determine the Range for \( \lambda \)
Since \( \sin^2 2x \) must be between 0 and 1:
\[ 0 \leq \frac{4(1 - \lambda)}{3} \leq 1 \]

### Step 8: Solve the Inequalities
1. From \( \frac{4(1 - \lambda)}{3} \geq 0 \):
\[ 1 - \lambda \geq 0 \implies \lambda \leq 1 \]
2. From \( \frac{4(1 - \lambda)}{3} \leq 1 \):
\[ 4(1 - \lambda) \leq 3 \implies 1 - \lambda \leq \frac{3}{4} \implies \lambda \geq \frac{1}{4} \]

### Final Result
Combining these results, we find:
\[ \frac{1}{4} \leq \lambda \leq 1 \]

Thus, the equation \( \sin^6 x + \cos^6 x = \lambda \) has a solution if:
\[ \lambda \in \left[ \frac{1}{4}, 1 \right] \]