Statement -1: `"cos"^(7) x + "sin"^(4) x = 1` has only two nonzero solutions in the inteval `(-pi, pi)`
Statement-2: `"cos"^(5)x + "cos"^(2)x -2=0` is possible only when cos x = 1.

Options

A. Statement -1 is true, Statement-2 is true, Statement -2 is a correct explanation for Statement-1.
B. Statement -1 is True, Statement-2 is True, Statement -2 is not a correct explanation for Statement -1.
C. Statement-1 is True, Statement-2 is False.
D. Statement -1 is False, Statement-2 is True.

Explanation

To solve the given statements, we will analyze each statement step by step.

### Statement 1:
**Equation:** \( \cos^7 x + \sin^4 x = 1 \)

1. **Rewrite the equation using the Pythagorean identity:**
\[ \sin^2 x = 1 - \cos^2 x \]
Therefore, we can express \( \sin^4 x \) as:
\[ \sin^4 x = (\sin^2 x)^2 = (1 - \cos^2 x)^2 \]
Substituting this into the equation gives:
\[ \cos^7 x + (1 - \cos^2 x)^2 = 1 \]

2. **Expand \( (1 - \cos^2 x)^2 \):**
\[ (1 - \cos^2 x)^2 = 1 - 2\cos^2 x + \cos^4 x \]
So the equation becomes:
\[ \cos^7 x + 1 - 2\cos^2 x + \cos^4 x = 1 \]

3. **Simplify the equation:**
\[ \cos^7 x + \cos^4 x - 2\cos^2 x = 0 \]

4. **Factor out \( \cos^2 x \):**
\[ \cos^2 x (\cos^5 x + \cos^2 x - 2) = 0 \]
This gives us two cases to consider:
- Case 1: \( \cos^2 x = 0 \)
- Case 2: \( \cos^5 x + \cos^2 x - 2 = 0 \)

5. **Solve Case 1:**
- If \( \cos^2 x = 0 \), then \( \cos x = 0 \). This occurs at:
\[ x = \pm \frac{\pi}{2} \]

6. **Solve Case 2:**
- Rearranging gives:
\[ \cos^5 x + \cos^2 x - 2 = 0 \]
Let \( y = \cos^2 x \), then the equation becomes:
\[ y^{\frac{5}{2}} + y - 2 = 0 \]
This is a polynomial in \( y \). We can check for values of \( y \) between 0 and 1 (since \( y = \cos^2 x \)).

7. **Finding the roots of \( y^{\frac{5}{2}} + y - 2 = 0 \):**
- Testing \( y = 1 \):
\[ 1^{\frac{5}{2}} + 1 - 2 = 0 \quad \text{(this is a root)} \]
- Therefore, \( \cos^2 x = 1 \) implies \( \cos x = 1 \) which gives \( x = 0 \).

8. **Summary of solutions:**
- From Case 1: \( x = \pm \frac{\pi}{2} \)
- From Case 2: \( x = 0 \)

Thus, the non-zero solutions in the interval \( (-\pi, \pi) \) are \( x = \frac{\pi}{2} \) and \( x = -\frac{\pi}{2} \).

### Conclusion for Statement 1:
The statement is true as there are only two non-zero solutions in the interval \( (-\pi, \pi) \).

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### Statement 2:
**Equation:** \( \cos^5 x + \cos^2 x - 2 = 0 \)

1. **Rearranging the equation:**
\[ \cos^5 x + \cos^2 x - 2 = 0 \]
This implies:
\[ \cos^5 x = 2 - \cos^2 x \]

2. **Analyzing the equation:**
- The maximum value of \( \cos^5 x \) is 1 (when \( \cos x = 1 \)).
- The right-hand side \( 2 - \cos^2 x \) can be at most 2 (when \( \cos^2 x = 0 \)), but for \( \cos^5 x \) to equal \( 2 - \cos^2 x \), we need \( \cos^5 x \) to equal 1.

3. **Setting \( \cos x = 1 \):**
- This gives \( \cos^5 x = 1 \) and \( \cos^2 x = 1 \), which satisfies the equation.

### Conclusion for Statement 2:
The statement is true, and it is only possible when \( \cos x = 1 \).

### Final Conclusion:
Both statements are true, but Statement 2 does not provide a correct explanation for Statement 1.