The number of distinct roots of the equation `A"sin"^(3) x + B"cos"^(3)x +C = 0` no two of which differ by `2pi`, is

Options

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

Answer

To find the number of distinct roots of the equation \( A \sin^3 x + B \cos^3 x + C = 0 \), we will follow a systematic approach.

### Step-by-Step Solution:

1. **Parametric Representation**:
We can express \(\sin x\) and \(\cos x\) in terms of a parameter \(t\):
\[ \sin x = \frac{2t}{1+t^2}, \quad \cos x = \frac{1-t^2}{1+t^2} \]

2. **Substituting into the Equation**:
Substitute these expressions into the equation:
\[ A \left(\frac{2t}{1+t^2}\right)^3 + B \left(\frac{1-t^2}{1+t^2}\right)^3 + C = 0 \]
This simplifies to:
\[ A \frac{8t^3}{(1+t^2)^3} + B \frac{(1-t^2)^3}{(1+t^2)^3} + C = 0 \]

3. **Clearing the Denominator**:
Multiply through by \((1+t^2)^3\) to eliminate the denominator:
\[ 8At^3 + B(1-t^2)^3 + C(1+t^2)^3 = 0 \]

4. **Expanding the Terms**:
Expand \(B(1-t^2)^3\) and \(C(1+t^2)^3\):
- \(B(1-t^2)^3 = B(1 - 3t^2 + 3t^4 - t^6)\)
- \(C(1+t^2)^3 = C(1 + 3t^2 + 3t^4 + t^6)\)

5. **Combining Like Terms**:
Combine all terms:
\[ 8At^3 + B(1 - 3t^2 + 3t^4 - t^6) + C(1 + 3t^2 + 3t^4 + t^6) = 0 \]
This gives us a polynomial in \(t\):
\[ (-B + C)t^6 + (3B + 3C)t^4 + (8A - 3B + 3C)t^3 + (B + C) = 0 \]

6. **Analyzing the Polynomial**:
The polynomial is of degree 6, which means it can have at most 6 roots.

7. **Distinct Roots**:
Since we are looking for distinct roots that do not differ by \(2\pi\), we need to consider the nature of the roots:
- The maximum number of distinct roots is 6, provided that the coefficients \(A\), \(B\), and \(C\) are such that all roots are distinct and real.

8. **Conclusion**:
Therefore, the number of distinct roots of the equation \( A \sin^3 x + B \cos^3 x + C = 0 \) is at most 6.

### Final Answer:
The number of distinct roots of the equation is **6**.