Find all possible triplets (x,y,z) such that `(x+y)+(y+2z) cos 2 theta +(z-x) sin^(2) theta =0` , for all ` theta `.

Answer

To solve the equation \((x+y) + (y + 2z) \cos 2\theta + (z - x) \sin^2 \theta = 0\) for all \(\theta\), we will analyze the equation step-by-step.

### Step 1: Rewrite the equation
We start with the equation:
\[ (x+y) + (y + 2z) \cos 2\theta + (z - x) \sin^2 \theta = 0 \]
This can be rearranged as:
\[ x + y + (y + 2z) \cos 2\theta + (z - x) \sin^2 \theta = 0 \]

### Step 2: Use the identity for \(\sin^2 \theta\)
We know that:
\[ \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \]
Substituting this into the equation gives:
\[ x + y + (y + 2z) \cos 2\theta + (z - x) \left(\frac{1 - \cos 2\theta}{2}\right) = 0 \]
Multiplying through by 2 to eliminate the fraction:
\[ 2(x + y) + 2(y + 2z) \cos 2\theta + (z - x)(1 - \cos 2\theta) = 0 \]

### Step 3: Expand the equation
Expanding the equation results in:
\[ 2x + 2y + 2(y + 2z) \cos 2\theta + (z - x) - (z - x) \cos 2\theta = 0 \]
This simplifies to:
\[ (2x + 2y + z - x) + (2y + 2z - z + x) \cos 2\theta = 0 \]
Rearranging terms gives:
\[ (2x + 2y + z - x) + (y + z + x) \cos 2\theta = 0 \]

### Step 4: Separate constant and variable terms
For this equation to hold for all \(\theta\), both the constant term and the coefficient of \(\cos 2\theta\) must independently equal zero:
1. Constant term:
\[ 2x + 2y + z - x = 0 \implies x + 2y + z = 0 \quad \text{(1)} \]
2. Coefficient of \(\cos 2\theta\):
\[ y + z + x = 0 \quad \text{(2)} \]

### Step 5: Solve the system of equations
From equation (2):
\[ y + z + x = 0 \implies z = -x - y \quad \text{(3)} \]
Substituting equation (3) into equation (1):
\[ x + 2y + (-x - y) = 0 \implies x + 2y - x - y = 0 \implies y = 0 \]
Substituting \(y = 0\) back into equation (3):
\[ z = -x - 0 = -x \]

### Step 6: Write the final triplet
Thus, we have:
\[ y = 0, \quad z = -x \]
The triplet can be expressed as:
\[ (x, 0, -x) \quad \text{for any real number } x. \]

### Final Answer:
The possible triplets \((x, y, z)\) are:
\[ (x, 0, -x) \quad \text{where } x \in \mathbb{R}. \]