Let's find the range of the expression:

\[ f(\theta) = 2\cos^2(\theta) - 6\sin(\theta) + 1 \]

### Step 1: Express \( \cos^2(\theta) \) in terms of \( \sin(\theta) \)

Using the Pythagorean identity:
\[ \cos^2(\theta) = 1 - \sin^2(\theta) \]

Substitute into the original function:
\[
f(\theta) = 2(1 - \sin^2(\theta)) - 6\sin(\theta) + 1
\]

Simplify this expression:
\[
f(\theta) = 2 - 2\sin^2(\theta) - 6\sin(\theta) + 1
\]

\[
f(\theta) = -2\sin^2(\theta) - 6\sin(\theta) + 3
\]

### Step 2: Analyze the quadratic expression in \( \sin(\theta) \)

This is now a quadratic function in \( \sin(\theta) \):
\[
f(x) = -2x^2 - 6x + 3
\]
where \( x = \sin(\theta) \) and \( x \in [-1, 1] \).

### Step 3: Find the vertex of the quadratic

The quadratic function \( f(x) = ax^2 + bx + c \) reaches its maximum or minimum value at:
\[
x = -\frac{b}{2a}
\]
Here, \( a = -2 \), \( b = -6 \).

So,
\[
x = -\frac{-6}{2(-2)} = \frac{6}{-4} = -\frac{3}{2}
\]

Since \( -\frac{3}{2} \) is outside the range \( [-1, 1] \) of \( \sin(\theta) \), we need to evaluate \( f(x) \) at the boundaries \( x = -1 \) and \( x = 1 \).

### Step 4: Evaluate the function at the boundaries

1. **At \( x = -1 \):**
   \[
   f(-1) = -2(-1)^2 - 6(-1) + 3 = -2(1) + 6 + 3 = 7
   \]

2. **At \( x = 1 \):**
   \[
   f(1) = -2(1)^2 - 6(1) + 3 = -2(1) - 6 + 3 = -5
   \]

### Conclusion
The maximum value is \( 7 \) and the minimum value is \( -5 \). 

So, the range of \( f(\theta) = 2\cos^2(\theta) - 6\sin(\theta) + 1 \) is \(\boxed{[-5, 7]}\).

Would you like further details or have any questions?

Here are 8 related questions you might find interesting:
1. What is the range of the function \( 3\sin^2(\theta) + 4\cos(\theta) \)?
2. How does the coefficient of \( \sin(\theta) \) affect the range of a quadratic trigonometric function?
3. What is the effect of squaring a trigonometric function on its range?
4. How can you find the maximum or minimum value of a quadratic function?
5. What role does the vertex play in determining the range of a quadratic function?
6. How does changing the constant term in a quadratic function affect its range?
7. Can a quadratic trigonometric function have only a minimum or maximum within the given domain?
8. What is the significance of evaluating a function at boundary points?

**Tip:** For quadratic trigonometric functions, always consider both the vertex and the boundary values when determining the range, especially when the function's vertex lies outside the typical range of \( \sin(\theta) \) or \( \cos(\theta) \).

Explore how to find the range of the function f(θ) = 2cos^2(θ) - 6sin(θ) + 1 using trigonometric identities and quadratic function analysis. This detailed solution covers the quadratic nature of the function and provides insights into trigonometric transformations. Suitable for high school students studying trigonometry and quadratic functions.

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