The number of integral triplets (a, b, c) such that `a+b cos 2x+c sin^(2)x=0` for all x, is

Options

A. 0
B. 1
C. 3
D. infinitelt many

Explanation

To solve the problem of finding the number of integral triplets \((a, b, c)\) such that the equation \(a + b \cos 2x + c \sin^2 x = 0\) holds for all \(x\), we can follow these steps:

### Step-by-Step Solution:

1. **Write the given equation**:
\[ a + b \cos 2x + c \sin^2 x = 0 \]

2. **Use the identity for \(\cos 2x\)**:
We know that \(\cos 2x = 1 - 2\sin^2 x\). Substitute this into the equation:
\[ a + b(1 - 2\sin^2 x) + c \sin^2 x = 0 \]

3. **Expand the equation**:
\[ a + b - 2b \sin^2 x + c \sin^2 x = 0 \]

4. **Combine like terms**:
Group the constant terms and the terms involving \(\sin^2 x\):
\[ (a + b) + (-2b + c) \sin^2 x = 0 \]

5. **Set coefficients to zero**:
For the equation to hold for all \(x\), both coefficients must be zero:
\[ a + b = 0 \quad \text{(1)} \]
\[ -2b + c = 0 \quad \text{(2)} \]

6. **Solve the equations**:
From equation (1):
\[ a = -b \]
From equation (2):
\[ c = 2b \]

7. **Express the triplet**:
Now we can express the triplet \((a, b, c)\) in terms of \(b\):
\[ (a, b, c) = (-b, b, 2b) \]

8. **Determine the integral triplets**:
Since \(b\) can take any integer value, we can conclude that there are infinitely many integral triplets \((a, b, c)\) that satisfy the equation.

### Conclusion:
Thus, the number of integral triplets \((a, b, c)\) such that \(a + b \cos 2x + c \sin^2 x = 0\) for all \(x\) is **infinitely many**.