If `sin x+sin^2=1`, then `cos^8 x+2 cos^6 x+cos^4 x=`

Options

A. 2
B. 1
C. 3
D. `(1)/(2)`

Answer

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

### Step 1: Rearranging the Equation
We start with the equation:
\[ \sin x + \sin^2 x = 1 \]
We can rearrange this to isolate \( \sin^2 x \):
\[ \sin^2 x + \sin x - 1 = 0 \]

### 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 \( a = 1, b = 1, c = -1 \):
\[ \sin x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ \sin x = \frac{-1 \pm \sqrt{1 + 4}}{2} \]
\[ \sin x = \frac{-1 \pm \sqrt{5}}{2} \]

### Step 3: Finding Valid Solutions
Since \( \sin x \) must be between -1 and 1, we need to check the two possible solutions:
1. \( \sin x = \frac{-1 + \sqrt{5}}{2} \) (valid)
2. \( \sin x = \frac{-1 - \sqrt{5}}{2} \) (not valid as it is less than -1)

Thus, we have:
\[ \sin x = \frac{-1 + \sqrt{5}}{2} \]

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

### Step 5: Finding \( \cos^4 x \)
Now, we calculate \( \cos^4 x \):
\[ \cos^4 x = \left(\cos^2 x\right)^2 = \left(\frac{-1 + \sqrt{5}}{2}\right)^2 = \frac{(-1 + \sqrt{5})^2}{4} = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2} \]

### Step 6: Finding \( \cos^6 x \) and \( \cos^8 x \)
Next, we find \( \cos^6 x \) and \( \cos^8 x \):
\[ \cos^6 x = \cos^4 x \cdot \cos^2 x = \left(\frac{3 - \sqrt{5}}{2}\right) \cdot \left(\frac{-1 + \sqrt{5}}{2}\right) \]
Calculating this gives:
\[ \cos^6 x = \frac{(3 - \sqrt{5})(-1 + \sqrt{5})}{4} = \frac{-3 + 3\sqrt{5} + \sqrt{5} - 5}{4} = \frac{-8 + 4\sqrt{5}}{4} = -2 + \sqrt{5} \]

Now for \( \cos^8 x \):
\[ \cos^8 x = \cos^4 x \cdot \cos^4 x = \left(\frac{3 - \sqrt{5}}{2}\right)^2 = \frac{(3 - \sqrt{5})^2}{4} = \frac{9 - 6\sqrt{5} + 5}{4} = \frac{14 - 6\sqrt{5}}{4} = \frac{7 - 3\sqrt{5}}{2} \]

### Step 7: Putting It All Together
Now we can substitute back into the expression:
\[ \cos^8 x + 2 \cos^6 x + \cos^4 x = \frac{7 - 3\sqrt{5}}{2} + 2(-2 + \sqrt{5}) + \frac{3 - \sqrt{5}}{2} \]
Calculating this gives:
\[ = \frac{7 - 3\sqrt{5}}{2} + \frac{-4 + 2\sqrt{5}}{1} + \frac{3 - \sqrt{5}}{2} \]
Combining like terms leads to:
\[ = \frac{7 - 3\sqrt{5} - 8 + 4\sqrt{5} + 3 - \sqrt{5}}{2} = \frac{2}{2} = 1 \]

### Final Answer
Thus, the final answer is:
\[ \cos^8 x + 2 \cos^6 x + \cos^4 x = 1 \]