If `sintheta+sin^(2)theta+sin^(3)theta=1`, then find the value of `cos^(6)theta-4cos^(4)theta+8cos^(2)theta`.

Options

A. 0
B. 3
C. 1
D. 4

Answer

To solve the equation \( \sin \theta + \sin^2 \theta + \sin^3 \theta = 1 \) and find the value of \( \cos^6 \theta - 4 \cos^4 \theta + 8 \cos^2 \theta \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \sin \theta + \sin^2 \theta + \sin^3 \theta = 1 \]
Let's denote \( x = \sin \theta \). Thus, we can rewrite the equation as:
\[ x + x^2 + x^3 = 1 \]

### Step 2: Rearrange the equation
Rearranging gives us:
\[ x^3 + x^2 + x - 1 = 0 \]

### Step 3: Factor the polynomial
To factor the polynomial, we can test for rational roots. By substituting \( x = 1 \):
\[ 1^3 + 1^2 + 1 - 1 = 1 + 1 + 1 - 1 = 2 \quad (\text{not a root}) \]
Now, let's try \( x = 0 \):
\[ 0^3 + 0^2 + 0 - 1 = -1 \quad (\text{not a root}) \]
Next, let's try \( x = -1 \):
\[ (-1)^3 + (-1)^2 + (-1) - 1 = -1 + 1 - 1 - 1 = -2 \quad (\text{not a root}) \]
Next, we can try \( x = \frac{1}{2} \):
\[ \left(\frac{1}{2}\right)^3 + \left(\frac{1}{2}\right)^2 + \frac{1}{2} - 1 = \frac{1}{8} + \frac{1}{4} + \frac{1}{2} - 1 = \frac{1}{8} + \frac{2}{8} + \frac{4}{8} - \frac{8}{8} = 0 \]
Thus, \( x = \frac{1}{2} \) is a root.

### Step 4: Factor out \( (x - \frac{1}{2}) \)
Now we can factor \( x^3 + x^2 + x - 1 \) as:
\[ (x - \frac{1}{2})(Ax^2 + Bx + C) \]
Using synthetic division or polynomial long division, we find:
\[ x^3 + x^2 + x - 1 = (x - \frac{1}{2})(x^2 + \frac{3}{2}x + 2) \]

### Step 5: Solve for \( x \)
The quadratic \( x^2 + \frac{3}{2}x + 2 = 0 \) has no real roots (discriminant is negative). Thus, the only real solution is:
\[ \sin \theta = \frac{1}{2} \]

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

### Step 7: Substitute \( \cos^2 \theta \) into the expression
Let \( y = \cos^2 \theta \). Then, \( y = \frac{3}{4} \). We need to find:
\[ \cos^6 \theta - 4 \cos^4 \theta + 8 \cos^2 \theta = y^3 - 4y^2 + 8y \]
Substituting \( y = \frac{3}{4} \):
\[ \left(\frac{3}{4}\right)^3 - 4\left(\frac{3}{4}\right)^2 + 8\left(\frac{3}{4}\right) \]

### Step 8: Calculate each term
Calculating each term:
1. \( \left(\frac{3}{4}\right)^3 = \frac{27}{64} \)
2. \( 4\left(\frac{3}{4}\right)^2 = 4 \cdot \frac{9}{16} = \frac{36}{16} = \frac{144}{64} \)
3. \( 8\left(\frac{3}{4}\right) = 6 \)

Now convert \( 6 \) to a fraction with a denominator of \( 64 \):
\[ 6 = \frac{384}{64} \]

### Step 9: Combine all terms
Now combine:
\[ \frac{27}{64} - \frac{144}{64} + \frac{384}{64} = \frac{27 - 144 + 384}{64} = \frac{267}{64} \]

### Final Answer
Thus, the value of \( \cos^6 \theta - 4 \cos^4 \theta + 8 \cos^2 \theta \) is:
\[ \frac{267}{64} \]