If `-pi < theta < pi`, the equation
`(cos 3 theta + 1)x^(2) + 2(cos 2theta - 1)x` `+(1 - 2 cos theta) = 0`
has more than two roots for

Options

A. no value of `theta`
B. one value of `theta`
C. two value of `theta`
D. all value of `theta`

Answer

To solve the equation

\[ (\cos 3\theta + 1)x^2 + 2(\cos 2\theta - 1)x + (1 - 2\cos \theta) = 0 \]

for the condition that it has more than two roots, we need to analyze the coefficients of the quadratic equation. A quadratic equation can have more than two roots if all its coefficients are zero.

### Step 1: Set the coefficients to zero

The coefficients of the quadratic equation are:

- \( a = \cos 3\theta + 1 \)
- \( b = 2(\cos 2\theta - 1) \)
- \( c = 1 - 2\cos \theta \)

For the equation to have more than two roots, we need:

1. \( \cos 3\theta + 1 = 0 \)
2. \( 2(\cos 2\theta - 1) = 0 \)
3. \( 1 - 2\cos \theta = 0 \)

### Step 2: Solve each equation

**1. Solve \( \cos 3\theta + 1 = 0 \)**

\[ \cos 3\theta = -1 \]

The cosine function equals -1 at odd multiples of \(\pi\):

\[ 3\theta = (2n + 1)\pi \quad \text{for } n \in \mathbb{Z} \]

Thus,

\[ \theta = \frac{(2n + 1)\pi}{3} \]

Considering the range \(-\pi < \theta < \pi\), we find:

- For \(n = -1\): \(\theta = -\frac{\pi}{3}\)
- For \(n = 0\): \(\theta = \frac{\pi}{3}\)

**2. Solve \( 2(\cos 2\theta - 1) = 0 \)**

\[ \cos 2\theta - 1 = 0 \implies \cos 2\theta = 1 \]

The cosine function equals 1 at multiples of \(2\pi\):

\[ 2\theta = 2n\pi \quad \text{for } n \in \mathbb{Z} \]

Thus,

\[ \theta = n\pi \]

Considering the range \(-\pi < \theta < \pi\), we find:

- For \(n = 0\): \(\theta = 0\)

**3. Solve \( 1 - 2\cos \theta = 0 \)**

\[ 2\cos \theta = 1 \implies \cos \theta = \frac{1}{2} \]

The cosine function equals \(\frac{1}{2}\) at:

\[ \theta = \frac{\pi}{3}, -\frac{\pi}{3} \]

### Step 3: Find the intersection of solutions

Now we have the solutions:

- From \( \cos 3\theta + 1 = 0 \): \(\theta = -\frac{\pi}{3}, \frac{\pi}{3}\)
- From \( 2(\cos 2\theta - 1) = 0 \): \(\theta = 0\)
- From \( 1 - 2\cos \theta = 0 \): \(\theta = -\frac{\pi}{3}, \frac{\pi}{3}\)

The intersection of these sets is:

- \(\theta = -\frac{\pi}{3}\) and \(\theta = \frac{\pi}{3}\) from the first and third equations, but \(0\) does not intersect.

### Conclusion

Since there are no common values satisfying all three equations, the solution indicates that there are no values of \(\theta\) for which the original equation has more than two roots.

### Final Answer

Thus, the answer is:

**There are no values of \(\theta\) for which the equation has more than two roots.**

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