Consider the cubic equation `x^3-(1+cos theta+sin theta)x^2+(cos theta sin theta+cos theta+sin theta)x-sin theta. cos theta =0` Whose roots are `x_1, x_2 and x_3`

Options

A. 3
B. 4
C. 5
D. 6

Explanation

To solve the cubic equation

\[ x^3 - (1 + \cos \theta + \sin \theta)x^2 + (\cos \theta \sin \theta + \cos \theta + \sin \theta)x - \sin \theta \cos \theta = 0 \]

whose roots are \( x_1, x_2, \) and \( x_3 \), we will use Vieta's formulas, which relate the coefficients of the polynomial to sums and products of its roots.

### Step 1: Identify coefficients
The given cubic equation can be compared to the general form of a cubic equation:

\[ ax^3 + bx^2 + cx + d = 0 \]

From the equation, we identify:
- \( a = 1 \)
- \( b = -(1 + \cos \theta + \sin \theta) \)
- \( c = \cos \theta \sin \theta + \cos \theta + \sin \theta \)
- \( d = -\sin \theta \cos \theta \)

### Step 2: Apply Vieta's Formulas
According to Vieta's formulas for a cubic equation:
1. The sum of the roots \( x_1 + x_2 + x_3 = -\frac{b}{a} \)
2. The sum of the product of the roots taken two at a time \( x_1 x_2 + x_2 x_3 + x_3 x_1 = \frac{c}{a} \)
3. The product of the roots \( x_1 x_2 x_3 = -\frac{d}{a} \)

Using these formulas, we can find the necessary values.

### Step 3: Calculate the sum of the roots
Using the first formula:
\[ x_1 + x_2 + x_3 = -\frac{-(1 + \cos \theta + \sin \theta)}{1} = 1 + \cos \theta + \sin \theta \]

### Step 4: Calculate the sum of the product of the roots taken two at a time
Using the second formula:
\[ x_1 x_2 + x_2 x_3 + x_3 x_1 = \frac{\cos \theta \sin \theta + \cos \theta + \sin \theta}{1} = \cos \theta \sin \theta + \cos \theta + \sin \theta \]

### Step 5: Calculate the product of the roots
Using the third formula:
\[ x_1 x_2 x_3 = -\frac{-\sin \theta \cos \theta}{1} = \sin \theta \cos \theta \]

### Step 6: Find \( x_1^2 + x_2^2 + x_3^2 \)
We can use the identity:
\[ x_1^2 + x_2^2 + x_3^2 = (x_1 + x_2 + x_3)^2 - 2(x_1 x_2 + x_2 x_3 + x_3 x_1) \]

Substituting the values we found:
1. \( (x_1 + x_2 + x_3)^2 = (1 + \cos \theta + \sin \theta)^2 \)
2. \( 2(x_1 x_2 + x_2 x_3 + x_3 x_1) = 2(\cos \theta \sin \theta + \cos \theta + \sin \theta) \)

Calculating \( (1 + \cos \theta + \sin \theta)^2 \):
\[ = 1 + 2(\cos \theta + \sin \theta) + (\cos^2 \theta + \sin^2 \theta) = 1 + 2(\cos \theta + \sin \theta) + 1 = 2 + 2(\cos \theta + \sin \theta) \]

Now substituting back:
\[ x_1^2 + x_2^2 + x_3^2 = (2 + 2(\cos \theta + \sin \theta)) - 2(\cos \theta \sin \theta + \cos \theta + \sin \theta) \]
\[ = 2 + 2(\cos \theta + \sin \theta) - 2\cos \theta \sin \theta - 2(\cos \theta + \sin \theta) \]
\[ = 2 - 2\cos \theta \sin \theta \]

### Final Result
Thus, the final expression for \( x_1^2 + x_2^2 + x_3^2 \) is:
\[ x_1^2 + x_2^2 + x_3^2 = 2 - 2\cos \theta \sin \theta \]