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

Answer

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

### Step 1: Simplify the given equation
We start with the equation:
\[ \sin \theta + \sin^2 \theta = 1 \]
Rearranging gives:
\[ \sin^2 \theta + \sin \theta - 1 = 0 \]
This is a quadratic equation in terms of \( \sin \theta \).

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

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

### Step 4: Substitute \( \cos^2 \theta \) into the expression
Let \( x = \cos^2 \theta \), then:
\[ x = \frac{-1 + \sqrt{5}}{2} \]
We need to evaluate:
\[ x^6 + 3x^5 + 3x^4 + x^3 - 1 \]

### Step 5: Calculate powers of \( x \)
Using \( x = \frac{-1 + \sqrt{5}}{2} \):
1. Calculate \( x^3, x^4, x^5, x^6 \):
- \( x^3 = x \cdot x^2 \)
- \( x^4 = x \cdot x^3 \)
- \( x^5 = x \cdot x^4 \)
- \( x^6 = x \cdot x^5 \)

### Step 6: Use the polynomial identity
Notice that \( x^2 + x - 1 = 0 \) implies \( x^2 = 1 - x \). We can express higher powers of \( x \) in terms of \( x \) and constants.

### Step 7: Substitute back into the expression
After calculating \( x^3, x^4, x^5, x^6 \), substitute them back into the expression:
\[ x^6 + 3x^5 + 3x^4 + x^3 - 1 = 0 \]

### Conclusion
Thus, the value of \( \cos^{12} \theta + 3\cos^{10} \theta + 3\cos^8 \theta + \cos^6 \theta - 1 \) is:
\[ \boxed{0} \]