`alpha ` is a root of equation `( 2 sin x - cos x ) (1+ cos x)=sin^2 x , beta ` is a root of the equation ` 3 cos ^2x - 10 cos x +3 =0 and gamma ` is a root of the equation `1-sin2 x = cos x- sin x : 0 le alpha , beta, gamma , le pi//2 ` `sin alpha + sin beta + sin gamma ` can be equal to

Options

A. `(14-3sqrt(2))/( 6 sqrt(2))`
B. `5//6`
C. `(3+4sqrt(2))/(6)`
D. `(1+sqrt(2))/(2) `

Answer

To solve the problem step by step, we will analyze each equation one by one and find the values of \( \sin \alpha \), \( \sin \beta \), and \( \sin \gamma \).

### Step 1: Solve for \( \alpha \)

The first equation is:
\[ (2 \sin x - \cos x)(1 + \cos x) = \sin^2 x \]

We can rewrite \( \sin^2 x \) using the identity \( \sin^2 x = 1 - \cos^2 x \):
\[ (2 \sin x - \cos x)(1 + \cos x) = 1 - \cos^2 x \]

Expanding the left side:
\[ (2 \sin x - \cos x)(1 + \cos x) = 2 \sin x + 2 \sin x \cos x - \cos x - \cos^2 x \]

Setting the equation:
\[ 2 \sin x + 2 \sin x \cos x - \cos x - \cos^2 x = 1 - \cos^2 x \]

Rearranging gives:
\[ 2 \sin x + 2 \sin x \cos x - \cos x = 1 \]

Factoring out \( (1 + \cos x) \):
\[ (1 + \cos x)(2 \sin x - 1) = 0 \]

This gives us two cases:
1. \( 1 + \cos x = 0 \) → \( \cos x = -1 \) (not possible in \( [0, \frac{\pi}{2}] \))
2. \( 2 \sin x - 1 = 0 \) → \( \sin x = \frac{1}{2} \)

Thus, \( \alpha = \frac{\pi}{6} \) and:
\[ \sin \alpha = \frac{1}{2} \]

### Step 2: Solve for \( \beta \)

The second equation is:
\[ 3 \cos^2 x - 10 \cos x + 3 = 0 \]

Let \( y = \cos x \), then we have:
\[ 3y^2 - 10y + 3 = 0 \]

Using the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 3 \cdot 3}}{2 \cdot 3} \]
\[ = \frac{10 \pm \sqrt{100 - 36}}{6} = \frac{10 \pm \sqrt{64}}{6} = \frac{10 \pm 8}{6} \]

This gives:
1. \( y = \frac{18}{6} = 3 \) (not possible)
2. \( y = \frac{2}{6} = \frac{1}{3} \)

Thus, \( \cos \beta = \frac{1}{3} \). Now, we find \( \sin \beta \):
\[ \sin^2 \beta + \cos^2 \beta = 1 \implies \sin^2 \beta = 1 - \left(\frac{1}{3}\right)^2 = 1 - \frac{1}{9} = \frac{8}{9} \]
\[ \sin \beta = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3} \]

### Step 3: Solve for \( \gamma \)

The third equation is:
\[ 1 - \sin 2x = \cos x - \sin x \]

Using the identity \( \sin 2x = 2 \sin x \cos x \):
\[ 1 - 2 \sin x \cos x = \cos x - \sin x \]

Rearranging gives:
\[ 2 \sin x \cos x + \sin x - \cos x + 1 = 0 \]

Factoring out \( \sin x \):
\[ \sin x (2 \cos x + 1) - \cos x + 1 = 0 \]

Setting \( \sin x = 0 \) gives \( \gamma = 0 \):
\[ \sin \gamma = 0 \]

### Step 4: Find the sum \( \sin \alpha + \sin \beta + \sin \gamma \)

Now we can find:
\[ \sin \alpha + \sin \beta + \sin \gamma = \frac{1}{2} + \frac{2\sqrt{2}}{3} + 0 \]

Finding a common denominator (which is 6):
\[ = \frac{3}{6} + \frac{4\sqrt{2}}{6} = \frac{3 + 4\sqrt{2}}{6} \]

### Final Answer
Thus, the value of \( \sin \alpha + \sin \beta + \sin \gamma \) is:
\[ \frac{3 + 4\sqrt{2}}{6} \]