(i) Is the equation `2 cos^(2)theta + cos theta` -6 =0 possible ?
(ii) Is the equation 2 `"sin"^(2)theta-costheta +4=0` can never be less than 2.

Answer

To solve the given questions step by step, we will address each part separately.

### Part (i): Is the equation \(2 \cos^2 \theta + \cos \theta - 6 = 0\) possible?

1. **Substitution**: Let \(t = \cos \theta\). Then the equation becomes:
\[ 2t^2 + t - 6 = 0 \]

2. **Factoring the Quadratic**: We can factor the quadratic equation:
\[ 2t^2 + 4t - 3t - 6 = 0 \]
This can be rearranged as:
\[ (2t^2 + 4t) + (-3t - 6) = 0 \]
Factoring gives:
\[ 2t(t + 2) - 3(t + 2) = 0 \]
Which simplifies to:
\[ (2t - 3)(t + 2) = 0 \]

3. **Finding Roots**: Setting each factor to zero gives:
\[ 2t - 3 = 0 \quad \Rightarrow \quad t = \frac{3}{2} \]
\[ t + 2 = 0 \quad \Rightarrow \quad t = -2 \]

4. **Checking Validity of Roots**: We know that \(t = \cos \theta\) must lie within the range \([-1, 1]\).
- \(t = \frac{3}{2}\) is outside this range.
- \(t = -2\) is also outside this range.

5. **Conclusion**: Since both roots are outside the range of \(\cos \theta\), the equation \(2 \cos^2 \theta + \cos \theta - 6 = 0\) is **not possible**.

### Part (ii): Is the equation \(2 \sin^2 \theta - \cos \theta + 4 = 0\) can never be less than 2?

1. **Rearranging the Equation**: We rewrite the equation:
\[ 2 \sin^2 \theta - \cos \theta + 4 = 0 \]

2. **Using the Pythagorean Identity**: We know that \(\sin^2 \theta = 1 - \cos^2 \theta\). Substitute this into the equation:
\[ 2(1 - \cos^2 \theta) - \cos \theta + 4 = 0 \]
This simplifies to:
\[ 2 - 2 \cos^2 \theta - \cos \theta + 4 = 0 \]
Which further simplifies to:
\[ -2 \cos^2 \theta - \cos \theta + 6 = 0 \]

3. **Rearranging Again**: Multiply through by -1 to make it easier to analyze:
\[ 2 \cos^2 \theta + \cos \theta - 6 = 0 \]

4. **Analyzing the Minimum Value**: From part (i), we found that the quadratic \(2 \cos^2 \theta + \cos \theta - 6 = 0\) does not have real roots, meaning it does not cross the x-axis. Therefore, it must either be entirely above or below the x-axis.

5. **Finding the Vertex**: The vertex of a quadratic \(ax^2 + bx + c\) occurs at \(x = -\frac{b}{2a}\):
\[ \cos \theta = -\frac{1}{2 \cdot 2} = -\frac{1}{4} \]
Substitute back to find the value:
\[ 2\left(-\frac{1}{4}\right)^2 + \left(-\frac{1}{4}\right) - 6 \]
This will yield a negative value, confirming that the quadratic is always above a certain value.

6. **Conclusion**: Since the quadratic does not cross the x-axis and has a minimum value greater than 2, the equation \(2 \sin^2 \theta - \cos \theta + 4 = 0\) can never be less than 2.