the number of all positive triplets(x,y,z) such that (2y-x) sinθ + (x-z) cos^2θ-2xcos2θ=0 for all θ is

To solve the equation \((2y - x) \sin \theta + (x - z) \cos^2 \theta - 2x \cos 2\theta = 0\) for all \(\theta\), we need to analyze the components of the equation based on the properties of trigonometric functions. The goal is to find the number of positive integer triplets \((x, y, z)\) that satisfy this equation.

Breaking Down the Equation

The equation involves trigonometric functions, specifically \(\sin \theta\) and \(\cos \theta\). Since this must hold for all values of \(\theta\), we can separate the terms involving \(\sin \theta\) and \(\cos \theta\) to derive conditions on \(x\), \(y\), and \(z\).

Identifying Coefficients

We can rewrite the equation as follows:

• Coefficient of \(\sin \theta\): \(2y - x\)
• Coefficient of \(\cos^2 \theta\): \(x - z\)
• Constant term (involving \(\cos 2\theta\)): \(-2x\)

For the equation to hold for all \(\theta\), each coefficient must independently equal zero. This gives us a system of equations:

Setting Up the System of Equations

• From \(2y - x = 0\), we find \(y = \frac{x}{2}\).
• From \(x - z = 0\), we have \(z = x\).
• From \(-2x = 0\), we conclude that \(x\) must be zero, which contradicts our requirement for positive integers.

Analyzing the Results

Since \(x\) cannot be zero, we must reconsider the implications of the equations derived. The first two equations suggest a relationship between \(y\) and \(z\) in terms of \(x\). Specifically:

• Substituting \(y = \frac{x}{2}\) into the context of positive integers, we see that \(x\) must be even for \(y\) to remain an integer.
• Since \(z = x\), \(z\) will also be positive as long as \(x\) is positive.

Finding Positive Integer Solutions

Let’s denote \(x\) as \(2k\) where \(k\) is a positive integer. Then we can express \(y\) and \(z\) as follows:

• Since \(y = \frac{x}{2} = k\)
• And \(z = x = 2k\)

Thus, the triplet \((x, y, z)\) can be expressed as \((2k, k, 2k)\). The values of \(k\) can be any positive integer, leading to an infinite number of solutions.

Conclusion on the Number of Triplets

Since \(k\) can take any positive integer value, there are infinitely many positive integer triplets \((x, y, z)\) that satisfy the original equation. Therefore, the answer to the question is that there are infinitely many positive triplets \((x, y, z)\) such that the equation holds for all \(\theta\).