If `sin^6 theta + cos^6 theta + k sin^2 2theta`=1 , then k is equal to :

Options

A. 1/2
B. 1/4
C. 4
D. 3/4

Answer

To solve the equation \( \sin^6 \theta + \cos^6 \theta + k \sin^2 2\theta = 1 \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ \sin^6 \theta + \cos^6 \theta + k \sin^2 2\theta = 1 \]

### Step 2: Use the identity for \( \sin^6 \theta + \cos^6 \theta \)
We can use the identity for the sum of cubes:
\[ A^3 + B^3 = (A + B)(A^2 - AB + B^2) \]
Let \( A = \sin^2 \theta \) and \( B = \cos^2 \theta \). Thus,
\[ \sin^6 \theta + \cos^6 \theta = (\sin^2 \theta + \cos^2 \theta)((\sin^2 \theta)^2 - \sin^2 \theta \cos^2 \theta + (\cos^2 \theta)^2) \]
Since \( \sin^2 \theta + \cos^2 \theta = 1 \), we have:
\[ \sin^6 \theta + \cos^6 \theta = 1 \cdot \left( \sin^4 \theta + \cos^4 \theta - \sin^2 \theta \cos^2 \theta \right) \]

### Step 3: Simplify \( \sin^4 \theta + \cos^4 \theta \)
Using the identity \( \sin^4 \theta + \cos^4 \theta = (\sin^2 \theta + \cos^2 \theta)^2 - 2\sin^2 \theta \cos^2 \theta \):
\[ \sin^4 \theta + \cos^4 \theta = 1 - 2\sin^2 \theta \cos^2 \theta \]
Thus,
\[ \sin^6 \theta + \cos^6 \theta = 1 - 2\sin^2 \theta \cos^2 \theta - \sin^2 \theta \cos^2 \theta = 1 - 3\sin^2 \theta \cos^2 \theta \]

### Step 4: Substitute back into the equation
Substituting this back into the original equation:
\[ 1 - 3\sin^2 \theta \cos^2 \theta + k \sin^2 2\theta = 1 \]

### Step 5: Use the identity for \( \sin^2 2\theta \)
Recall that:
\[ \sin^2 2\theta = 4\sin^2 \theta \cos^2 \theta \]
Substituting this into the equation gives:
\[ 1 - 3\sin^2 \theta \cos^2 \theta + k(4\sin^2 \theta \cos^2 \theta) = 1 \]

### Step 6: Simplify the equation
Cancelling \( 1 \) from both sides:
\[ -3\sin^2 \theta \cos^2 \theta + 4k\sin^2 \theta \cos^2 \theta = 0 \]
Factoring out \( \sin^2 \theta \cos^2 \theta \):
\[ \sin^2 \theta \cos^2 \theta (-3 + 4k) = 0 \]

### Step 7: Solve for \( k \)
Since \( \sin^2 \theta \cos^2 \theta \) cannot be zero for all \( \theta \), we have:
\[ -3 + 4k = 0 \implies 4k = 3 \implies k = \frac{3}{4} \]

### Final Answer
Thus, the value of \( k \) is:
\[ \boxed{\frac{3}{4}} \]