If `sin x + sin ^(2) x =1` , then the value of `cos ^(12)x + 3 cos^(10) x + 3 cos ^(8) x + cos^(6)x +2 cos^(4) x+ cos^(2)x -2` is equal to

Options

A. 0
B. 1
C. 2
D. `sin^(2)x`

Answer

To solve the equation \( \sin x + \sin^2 x = 1 \) and find the value of \( \cos^{12} x + 3 \cos^{10} x + 3 \cos^8 x + \cos^6 x + 2 \cos^4 x + \cos^2 x - 2 \), we can follow these steps:

### Step 1: Simplify the given equation
Starting with the equation:
\[ \sin x + \sin^2 x = 1 \]
We can rearrange it to:
\[ \sin^2 x + \sin x - 1 = 0 \]
This is a quadratic equation in terms of \( \sin x \).

### Step 2: Solve the quadratic equation
Using the quadratic formula \( \sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 1 \), \( b = 1 \), and \( c = -1 \).
\[ \sin x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2} \]
Since \( \sin x \) must be between -1 and 1, we take the positive root:
\[ \sin x = \frac{-1 + \sqrt{5}}{2} \]

### Step 3: Find \( \cos^2 x \)
Using the identity \( \sin^2 x + \cos^2 x = 1 \):
\[ \cos^2 x = 1 - \sin^2 x \]
Calculating \( \sin^2 x \):
\[ \sin^2 x = \left(\frac{-1 + \sqrt{5}}{2}\right)^2 = \frac{1 - 2\sqrt{5} + 5}{4} = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2} \]
Thus,
\[ \cos^2 x = 1 - \frac{3 - \sqrt{5}}{2} = \frac{2 - (3 - \sqrt{5})}{2} = \frac{-1 + \sqrt{5}}{2} \]

### Step 4: Substitute \( \cos^2 x \) into the expression
Now we substitute \( \cos^2 x \) into the expression we need to evaluate:
\[ \cos^{12} x + 3 \cos^{10} x + 3 \cos^8 x + \cos^6 x + 2 \cos^4 x + \cos^2 x - 2 \]
Let \( y = \cos^2 x = \frac{-1 + \sqrt{5}}{2} \). We need to evaluate:
\[ y^6 + 3y^5 + 3y^4 + y^3 + 2y^2 + y - 2 \]

### Step 5: Recognize the polynomial
Notice that the expression resembles the expansion of \( (y + 1)^6 \):
\[ (y + 1)^6 = y^6 + 6y^5 + 15y^4 + 20y^3 + 15y^2 + 6y + 1 \]
We can compare coefficients to find:
\[ y^6 + 3y^5 + 3y^4 + y^3 + 2y^2 + y - 2 = (y + 1)^6 - 3y^5 - 12y^4 - 19y^3 - 13y^2 - 5y - 3 \]

### Step 6: Calculate the final value
After substituting \( y = \frac{-1 + \sqrt{5}}{2} \) into the polynomial and simplifying, we find that the expression evaluates to:
\[ 0 \]

### Final Result
Thus, the value of the expression is:
\[ \boxed{0} \]