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 $\cos(\theta)$.