If `sinx+sin^2x=1`, then find the value of `cos^(12)x +3cos^(10)x + 3 cos^8x + cos^6x-2 `

Options

A. 0
B. 1
C. 2
D. `-1`

Answer

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

### Step 1: Rearranging the equation
We start with the equation:
\[ \sin x + \sin^2 x = 1 \]
Rearranging gives:
\[ \sin^2 x + \sin x - 1 = 0 \]

### Step 2: Solving the quadratic equation
This is a quadratic equation in terms of \( \sin x \). We can use the quadratic formula:
\[ \sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = 1, c = -1 \).
Substituting these values, we get:
\[ \sin x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2} \]

### Step 3: Finding valid values for \( \sin x \)
The possible values for \( \sin x \) are:
\[ \sin x = \frac{-1 + \sqrt{5}}{2} \quad \text{or} \quad \sin x = \frac{-1 - \sqrt{5}}{2} \]
Since \( \sin x \) must be in the range \([-1, 1]\), we only consider:
\[ \sin x = \frac{-1 + \sqrt{5}}{2} \]

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

### Step 5: Substituting into the expression
We need to find:
\[ \cos^{12} x + 3 \cos^{10} x + 3 \cos^8 x + \cos^6 x - 2 \]
Let \( y = \cos^2 x = \frac{-1 + \sqrt{5}}{2} \). We can rewrite the expression as:
\[ y^6 + 3y^5 + 3y^4 + y^3 - 2 \]

### Step 6: Using the identity
Notice that \( y^2 + y - 1 = 0 \) implies \( y^2 = 1 - y \). We can use this to express higher powers of \( y \):
- \( y^3 = y \cdot y^2 = y(1 - y) = y - y^2 = y - (1 - y) = 2y - 1 \)
- \( y^4 = y \cdot y^3 = y(2y - 1) = 2y^2 - y = 2(1 - y) - y = 2 - 3y \)
- \( y^5 = y \cdot y^4 = y(2 - 3y) = 2y - 3y^2 = 2y - 3(1 - y) = 5y - 3 \)
- \( y^6 = y \cdot y^5 = y(5y - 3) = 5y^2 - 3y = 5(1 - y) - 3y = 5 - 8y \)

### Step 7: Substituting back into the expression
Now substituting these into the expression:
\[ (5 - 8y) + 3(5y - 3) + 3(2 - 3y) + (2y - 1) - 2 \]
Simplifying this:
\[ 5 - 8y + 15y - 9 + 6 - 9y + 2y - 1 - 2 \]
Combining like terms:
\[ (5 - 9 + 6 - 1 - 2) + (-8y + 15y - 9y + 2y) = -1 + 0y = -1 \]

### Final Answer
Thus, the value of the expression is:
\[ \boxed{-1} \]