if ` sin x + sin^2 x = 1`, then the value of `cos^2 x + cos^4x` is

Options

A. 1
B. 2
C. `1.5`
D. none of these

Answer

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

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

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

### Step 3: Finding the value of \( \cos^2 x \)
Since \( \sin^2 x + \cos^2 x = 1 \), we can express \( \cos^2 x \) in terms of \( \sin x \):
\[ \cos^2 x = 1 - \sin^2 x \]
Using \( \sin x = \frac{-1 + \sqrt{5}}{2} \) (we take the positive root since sine values range from -1 to 1), we calculate \( \sin^2 x \):
\[ \sin^2 x = \left(\frac{-1 + \sqrt{5}}{2}\right)^2 = \frac{(-1 + \sqrt{5})^2}{4} = \frac{1 - 2\sqrt{5} + 5}{4} = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2} \]
Now substituting this into the equation for \( \cos^2 x \):
\[ \cos^2 x = 1 - \frac{3 - \sqrt{5}}{2} = \frac{2 - (3 - \sqrt{5})}{2} = \frac{-1 + \sqrt{5}}{2} \]

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

### Step 5: Adding \( \cos^2 x \) and \( \cos^4 x \)
Now we can find \( \cos^2 x + \cos^4 x \):
\[ \cos^2 x + \cos^4 x = \frac{-1 + \sqrt{5}}{2} + \frac{3 - \sqrt{5}}{2} = \frac{-1 + \sqrt{5} + 3 - \sqrt{5}}{2} = \frac{2}{2} = 1 \]

### Final Answer
Thus, the value of \( \cos^2 x + \cos^4 x \) is:
\[ \boxed{1} \]