The number of all possible triplets `(p, q, r)` such that `p + qcos2theta+ rsin^2 theta = 0` for all `theta`, is

Options

A. `(k,-k, -2k)`
B. `(k/2, -k/2, -k) `
C. `(-k/2, k/2, k)`
D. `(-k, k, 2k)`

Answer

To solve the problem of finding the number of all possible triplets \((p, q, r)\) such that

\[ p + q \cos(2\theta) + r \sin^2(\theta) = 0 \]

for all \(\theta\), we will follow these steps:

### Step 1: Rewrite \(\cos(2\theta)\)
Using the trigonometric identity, we can rewrite \(\cos(2\theta)\) as:

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

Substituting this into the equation gives:

\[ p + q(1 - 2\sin^2(\theta)) + r \sin^2(\theta) = 0 \]

### Step 2: Simplify the Equation
Distributing \(q\) in the equation:

\[ p + q - 2q \sin^2(\theta) + r \sin^2(\theta) = 0 \]

Rearranging, we have:

\[ (p + q) + (-2q + r) \sin^2(\theta) = 0 \]

### Step 3: Set Coefficients to Zero
For the equation to hold for all \(\theta\), both coefficients must equal zero:

1. \(p + q = 0\)
2. \(-2q + r = 0\)

### Step 4: Solve for \(p\), \(q\), and \(r\)
From the first equation:

\[ p = -q \]

From the second equation:

\[ r = 2q \]

### Step 5: Express Triplets in Terms of \(q\)
Now we can express the triplet \((p, q, r)\) in terms of \(q\):

\[ (p, q, r) = (-q, q, 2q) \]

Let \(q = k\) (where \(k\) is any real number), then:

\[ (p, q, r) = (-k, k, 2k) \]

### Step 6: Identify Ratios
The triplet can be expressed in the ratio:

\[ (-1, 1, 2) \]

### Step 7: Consider Different Values of \(q\)
We can also consider other values of \(q\):

1. If \(q = -k\), then:
- \(p = -(-k) = k\)
- \(r = 2(-k) = -2k\)
- Triplet: \((k, -k, -2k)\)

2. If \(q = \frac{k}{2}\), then:
- \(p = -\frac{k}{2}\)
- \(r = 2\left(\frac{k}{2}\right) = k\)
- Triplet: \(\left(-\frac{k}{2}, \frac{k}{2}, k\right)\)

### Conclusion
Thus, we have identified that the triplets can be represented in multiple forms based on the values of \(q\). The possible ratios of triplets are:

1. \((-1, 1, 2)\)
2. \((1, -1, -2)\)
3. \(\left(-\frac{1}{2}, \frac{1}{2}, 1\right)\)

Since we can choose any real number for \(k\), there are infinitely many triplets \((p, q, r)\) that satisfy the equation.

### Final Answer
The number of all possible triplets \((p, q, r)\) is infinite.