If `alpha gt (1)/(sin^(6)x+cos^(6)x) AA x in R`, then `alpha` can be

Options

A. 3
B. 4
C. 5
D. 6

Answer

To solve the problem, we need to find the possible values of \( \alpha \) such that

\[ \alpha > \frac{1}{\sin^6 x + \cos^6 x} \]

for all \( x \in \mathbb{R} \).

### Step 1: Rewrite \( \sin^6 x + \cos^6 x \)

We can express \( \sin^6 x + \cos^6 x \) using the identity for the sum of cubes:

\[ \sin^6 x + \cos^6 x = (\sin^2 x)^3 + (\cos^2 x)^3 \]

Using the identity \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \), where \( a = \sin^2 x \) and \( b = \cos^2 x \):

\[ \sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)((\sin^2 x)^2 - \sin^2 x \cos^2 x + (\cos^2 x)^2) \]

### Step 2: Simplify using \( \sin^2 x + \cos^2 x = 1 \)

Since \( \sin^2 x + \cos^2 x = 1 \), we have:

\[ \sin^6 x + \cos^6 x = 1 \cdot \left((\sin^2 x)^2 - \sin^2 x \cos^2 x + (\cos^2 x)^2\right) \]

### Step 3: Further simplify

Now, we can express \( (\sin^2 x)^2 + (\cos^2 x)^2 \):

\[ (\sin^2 x)^2 + (\cos^2 x)^2 = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - 2\sin^2 x \cos^2 x \]

Thus, we have:

\[ \sin^6 x + \cos^6 x = 1 - 3\sin^2 x \cos^2 x \]

### Step 4: Use the double angle identity

Using the identity \( \sin^2 x \cos^2 x = \frac{1}{4} \sin^2(2x) \):

\[ \sin^6 x + \cos^6 x = 1 - \frac{3}{4} \sin^2(2x) \]

### Step 5: Determine the minimum value of \( \sin^6 x + \cos^6 x \)

The maximum value of \( \sin^2(2x) \) is 1, so the minimum value of \( \sin^6 x + \cos^6 x \) occurs when \( \sin^2(2x) = 1 \):

\[ \sin^6 x + \cos^6 x \text{ (min)} = 1 - \frac{3}{4} = \frac{1}{4} \]

### Step 6: Find \( \alpha \)

Now substituting back into our inequality:

\[ \alpha > \frac{1}{\sin^6 x + \cos^6 x} \]

The minimum value of \( \sin^6 x + \cos^6 x \) is \( \frac{1}{4} \), thus:

\[ \alpha > \frac{1}{\frac{1}{4}} = 4 \]

### Conclusion

Therefore, the values of \( \alpha \) must be greater than 4.

### Final Answer

\[ \alpha > 4 \]