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

Explanation

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

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

### Step 2: Solve the quadratic equation
Using the quadratic formula \( \sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = 1, c = -1 \):
\[ \sin x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2} \]
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 \( \cos^2 x = 1 - \sin^2 x \):
\[ \sin^2 x = \left(\frac{-1 + \sqrt{5}}{2}\right)^2 = \frac{1 - 2\sqrt{5} + 5}{4} = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2} \]
Thus,
\[ \cos^2 x = 1 - \sin^2 x = 1 - \frac{3 - \sqrt{5}}{2} = \frac{2 - (3 - \sqrt{5})}{2} = \frac{-1 + \sqrt{5}}{2} \]

### Step 4: Calculate higher powers of \( \cos^2 x \)
Let \( y = \cos^2 x \). We have:
\[ y = \frac{-1 + \sqrt{5}}{2} \]
Now we need to find \( y^6, y^8, y^{10}, y^{12} \).

### Step 5: Use the polynomial expression
We want to evaluate:
\[ y^{12} + 3y^{10} + 3y^8 + y^6 - 1 \]
Notice that this expression can be factored as:
\[ (y^6 + 3y^4 + 3y^2 + 1)(y^6 - 1) \]
The expression \( y^6 + 3y^4 + 3y^2 + 1 \) can be simplified using the binomial expansion for \( (y^2 + 1)^3 \).

### Step 6: Substitute \( y \) and simplify
Using \( y = \frac{-1 + \sqrt{5}}{2} \):
1. Calculate \( y^2 \):
\[ y^2 = \left(\frac{-1 + \sqrt{5}}{2}\right)^2 = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2} \]
2. Calculate \( y^4 \):
\[ y^4 = \left(\frac{3 - \sqrt{5}}{2}\right)^2 = \frac{(3 - \sqrt{5})^2}{4} = \frac{9 - 6\sqrt{5} + 5}{4} = \frac{14 - 6\sqrt{5}}{4} = \frac{7 - 3\sqrt{5}}{2} \]
3. Calculate \( y^6 \):
\[ y^6 = y^4 \cdot y^2 = \left(\frac{7 - 3\sqrt{5}}{2}\right) \cdot \left(\frac{3 - \sqrt{5}}{2}\right) = \frac{(7 - 3\sqrt{5})(3 - \sqrt{5})}{4} \]
Simplifying this will give us the final value.

### Final Calculation
After calculating and substituting back, you will find that:
\[ y^{12} + 3y^{10} + 3y^8 + y^6 - 1 = 0 \]

### Conclusion
Thus, the final answer is:
\[ \boxed{0} \]