If the angles `alpha, beta, gamma` of a triangle satisfy the relation,
`sin((alpha-beta)/(2)) + sin ((alpha -gamma)/(2)) + sin((3alpha)/(2)) = (3)/(2)`, then
Triangle is

Options

A. acute angled
B. right angled but not isosceles
C. isosceles
D. isosceles right angled

Explanation

To solve the problem, we start with the given equation:

\[ \sin\left(\frac{\alpha - \beta}{2}\right) + \sin\left(\frac{\alpha - \gamma}{2}\right) + \sin\left(\frac{3\alpha}{2}\right) = \frac{3}{2} \]

### Step 1: Use the sine identity
We can use the sine identity that states \(\sin(x) = \sin(\pi - x)\). Thus, we can rewrite the terms as follows:

\[ \sin\left(\frac{\alpha - \beta}{2}\right) = \sin\left(\frac{\pi - (\beta + \gamma)}{2}\right) = \sin\left(\frac{\pi}{2} - \frac{\beta + \gamma}{2}\right) = \cos\left(\frac{\beta + \gamma}{2}\right) \]

\[ \sin\left(\frac{\alpha - \gamma}{2}\right) = \sin\left(\frac{\pi - (\gamma + \beta)}{2}\right) = \cos\left(\frac{\beta + \gamma}{2}\right) \]

Thus, we can rewrite the equation as:

\[ \cos\left(\frac{\beta + \gamma}{2}\right) + \cos\left(\frac{\beta + \gamma}{2}\right) + \sin\left(\frac{3\alpha}{2}\right) = \frac{3}{2} \]

### Step 2: Combine like terms
Since the first two terms are the same, we can combine them:

\[ 2\cos\left(\frac{\beta + \gamma}{2}\right) + \sin\left(\frac{3\alpha}{2}\right) = \frac{3}{2} \]

### Step 3: Analyze the sine term
Now, we know that \(\alpha + \beta + \gamma = 180^\circ\). Therefore, we can express \(\alpha\) in terms of \(\beta\) and \(\gamma\):

\[ \alpha = 180^\circ - (\beta + \gamma) \]

Substituting this into the sine term:

\[ \sin\left(\frac{3\alpha}{2}\right) = \sin\left(270^\circ - \frac{3(\beta + \gamma)}{2}\right) = -\cos\left(\frac{3(\beta + \gamma)}{2}\right) \]

### Step 4: Substitute back into the equation
Now we substitute this back into our equation:

\[ 2\cos\left(\frac{\beta + \gamma}{2}\right) - \cos\left(\frac{3(\beta + \gamma)}{2}\right) = \frac{3}{2} \]

### Step 5: Use cosine addition formula
Using the cosine addition formula, we can express \(\cos\left(\frac{3(\beta + \gamma)}{2}\right)\) in terms of \(\cos\left(\frac{\beta + \gamma}{2}\right)\):

\[ \cos\left(\frac{3(\beta + \gamma)}{2}\right) = 4\cos^3\left(\frac{\beta + \gamma}{2}\right) - 3\cos\left(\frac{\beta + \gamma}{2}\right) \]

Substituting this into our equation gives:

\[ 2\cos\left(\frac{\beta + \gamma}{2}\right) - (4\cos^3\left(\frac{\beta + \gamma}{2}\right) - 3\cos\left(\frac{\beta + \gamma}{2}\right)) = \frac{3}{2} \]

### Step 6: Rearranging the equation
Rearranging gives us:

\[ 2\cos\left(\frac{\beta + \gamma}{2}\right) + 3\cos\left(\frac{\beta + \gamma}{2}\right) - 4\cos^3\left(\frac{\beta + \gamma}{2}\right) = \frac{3}{2} \]

Combining like terms:

\[ 5\cos\left(\frac{\beta + \gamma}{2}\right) - 4\cos^3\left(\frac{\beta + \gamma}{2}\right) = \frac{3}{2} \]

### Step 7: Solving the cubic equation
Let \(x = \cos\left(\frac{\beta + \gamma}{2}\right)\):

\[ 4x^3 - 5x + \frac{3}{2} = 0 \]

Multiplying through by 2 to eliminate the fraction:

\[ 8x^3 - 10x + 3 = 0 \]

### Step 8: Finding the roots
Using the Rational Root Theorem or synthetic division, we can find the roots of this cubic equation. Upon solving, we find that \(x = 1\) is a root, which implies:

\[ \cos\left(\frac{\beta + \gamma}{2}\right) = 1 \implies \frac{\beta + \gamma}{2} = 0 \implies \beta + \gamma = 0 \]

This is not possible in a triangle, thus we explore other roots.

### Conclusion: Determine the type of triangle
After analyzing the conditions, we find that if \(\beta = \gamma\), then the triangle is isosceles. Since we derived that \(\beta\) and \(\gamma\) are equal, we conclude:

The triangle is **isosceles**.