`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 `
` cos alpha + cos beta + cos gamma ` can be equal to

Options

A. `(3sqrt(6)+2sqrt(2)+6)/(6sqrt(2))`
B. `(3sqrt(3)-8)/(6)`
C. `(3sqrt(3)+2)/(6)`
D. None of these

Answer

To solve the problem step by step, we need to find the values of \( \alpha \), \( \beta \), and \( \gamma \) from the given equations and then calculate \( \cos \alpha + \cos \beta + \cos \gamma \).

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

We start with the equation:
\[ (2 \sin x - \cos x)(1 + \cos x) = \sin^2 x \]

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

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

Using the identity \( \sin^2 x + \cos^2 x = 1 \):
\[ 2 \sin x + 2 \sin x \cos x - \cos x - 1 = 0 \]

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

Setting each factor to zero:
1. \( 1 + \cos x = 0 \) gives \( \cos x = -1 \) (not valid in \( [0, \frac{\pi}{2}] \)).
2. \( 2 \sin x - 1 = 0 \) gives \( \sin x = \frac{1}{2} \) which implies \( x = \frac{\pi}{6} \).

Thus, \( \alpha = \frac{\pi}{6} \).

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

Next, we solve the equation:
\[ 3 \cos^2 x - 10 \cos x + 3 = 0 \]

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

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

Calculating the roots:
1. \( y = \frac{18}{6} = 3 \) (not valid).
2. \( y = \frac{2}{6} = \frac{1}{3} \).

Thus, \( \cos \beta = \frac{1}{3} \) implies \( \beta = \cos^{-1}(\frac{1}{3}) \).

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

Now, we solve:
\[ 1 - \sin 2x = \cos x - \sin x \]

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

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

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

Setting each factor to zero:
1. \( 1 - \cos x = 0 \) gives \( \cos x = 1 \) (implies \( x = 0 \)).
2. \( 1 - 2 \cos x = 0 \) gives \( \cos x = \frac{1}{2} \) which implies \( x = \frac{\pi}{3} \).

Thus, \( \gamma = 0 \) or \( \frac{\pi}{3} \).

### Step 4: Calculate \( \cos \alpha + \cos \beta + \cos \gamma \)

Now we find:
- \( \cos \alpha = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2} \)
- \( \cos \beta = \frac{1}{3} \)
- \( \cos \gamma = 1 \) (if \( \gamma = 0 \)) or \( \cos \frac{\pi}{3} = \frac{1}{2} \).

Calculating \( \cos \alpha + \cos \beta + \cos \gamma \):
1. If \( \gamma = 0 \):
\[ \cos \alpha + \cos \beta + \cos \gamma = \frac{\sqrt{3}}{2} + \frac{1}{3} + 1 \]
Finding a common denominator (6):
\[ = \frac{3\sqrt{3}}{6} + \frac{2}{6} + \frac{6}{6} = \frac{3\sqrt{3} + 8}{6} \]

2. If \( \gamma = \frac{\pi}{3} \):
\[ \cos \alpha + \cos \beta + \cos \gamma = \frac{\sqrt{3}}{2} + \frac{1}{3} + \frac{1}{2} \]
Finding a common denominator (6):
\[ = \frac{3\sqrt{3}}{6} + \frac{2}{6} + \frac{3}{6} = \frac{3\sqrt{3} + 5}{6} \]

### Final Answer
Thus, the possible values for \( \cos \alpha + \cos \beta + \cos \gamma \) can be:
- \( \frac{3\sqrt{3} + 8}{6} \) or \( \frac{3\sqrt{3} + 5}{6} \).