Key Concepts and Formulas

• Trigonometric Identities:
  • $\cot^2 \theta = \frac{\cos^2 \theta}{\sin^2 \theta}$
  • $\cos^2 \theta = 1 - \sin^2 \theta$
  • $\sin(2\theta) = 2\sin\theta\cos\theta$
  • $\cos(2\theta) = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$
• Solving Trigonometric Equations: Transforming the equation into a more manageable form, often a polynomial in a single trigonometric function, and finding solutions within a specified interval.
• Definite Integration: Evaluating the integral of a function between two limits.
  • $\int \cos^2(ax) \, dx = \int \frac{1 + \cos(2ax)}{2} \, dx = \frac{1}{2}x + \frac{\sin(2ax)}{4a} + C$
  • The Fundamental Theorem of Calculus: $\int_a^b f(x) \, dx = F(b) - F(a)$, where $F'(x) = f(x)$.

Step-by-Step Solution

Step 1: Interpret and Simplify the Given Trigonometric Equation

The given equation is $2\cot^2 \theta - \frac{5}{\sin \theta} + 4 = 0$. We are looking for solutions in the interval $(0, 2\pi) - \{\pi\}$. The condition $\theta \neq \pi$ ensures that $\sin \theta \neq 0$, so the terms involving $\sin \theta$ and $\cot \theta$ are well-defined.

We can rewrite $\cot^2 \theta$ using the identity $\cot^2 \theta = \frac{\cos^2 \theta}{\sin^2 \theta}$. Substituting this into the equation: $2 \frac{\cos^2 \theta}{\sin^2 \theta} - \frac{5}{\sin \theta} + 4 = 0$ Now, replace $\cos^2 \theta$ with $1 - \sin^2 \theta$: $2 \frac{1 - \sin^2 \theta}{\sin^2 \theta} - \frac{5}{\sin \theta} + 4 = 0$ To simplify, let $y = \sin \theta$. The equation becomes: $2 \frac{1 - y^2}{y^2} - \frac{5}{y} + 4 = 0$ Multiply the entire equation by $y^2$ (since $\sin \theta \neq 0$, $y \neq 0$): $2(1 - y^2) - 5y + 4y^2 = 0$ $2 - 2y^2 - 5y + 4y^2 = 0$ Combine like terms to form a quadratic equation in $y$: $2y^2 - 5y + 2 = 0$

Step 2: Solve the Quadratic Equation for $y = \sin \theta$

We can solve the quadratic equation $2y^2 - 5y + 2 = 0$ by factoring or using the quadratic formula. Factoring the quadratic: We look for two numbers that multiply to $(2)(2) = 4$ and add up to $-5$. These numbers are $-1$ and $-4$. $2y^2 - 4y - y + 2 = 0$ $2y(y - 2) - 1(y - 2) = 0$ $(2y - 1)(y - 2) = 0$ This gives two possible values for $y$: $2y - 1 = 0 \implies y = \frac{1}{2}$ $y - 2 = 0 \implies y = 2$

Step 3: Find the Values of $\theta$ from the Possible Values of $\sin \theta$

We have $y = \sin \theta$. So, we have two cases: Case 1: $\sin \theta = \frac{1}{2}$ We need to find the values of $\theta$ in $(0, 2\pi) - \{\pi\}$ such that $\sin \theta = \frac{1}{2}$. The general solutions for $\sin \theta = \frac{1}{2}$ are $\theta = n\pi + (-1)^n \frac{\pi}{6}$, where $n$ is an integer. For $n=0$, $\theta = \frac{\pi}{6}$. This is in the interval $(0, 2\pi)$ and $\neq \pi$. For $n=1$, $\theta = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$. This is in the interval $(0, 2\pi)$ and $\neq \pi$. For $n=2$, $\theta = 2\pi + \frac{\pi}{6} = \frac{13\pi}{6}$, which is outside the interval. For $n=-1$, $\theta = -\pi - \frac{\pi}{6} = -\frac{7\pi}{6}$, which is outside the interval.

Case 2: $\sin \theta = 2$ The sine function has a maximum value of 1. Therefore, $\sin \theta = 2$ has no real solutions for $\theta$.

So, the only valid values of $\theta$ in the given domain are $\theta = \frac{\pi}{6}$ and $\theta = \frac{5\pi}{6}$.

Step 4: Identify the Smallest and Largest Values of $\theta$

The problem states that $\theta_1$ and $\theta_2$ are the smallest and largest values of $\theta$ in $(0, 2\pi) - \{\pi\}$ that satisfy the equation. From our solutions, the values are $\frac{\pi}{6}$ and $\frac{5\pi}{6}$. The smallest value is $\theta_1 = \frac{\pi}{6}$. The largest value is $\theta_2 = \frac{5\pi}{6}$.

Step 5: Evaluate the Definite Integral

We need to calculate the definite integral: $I = \int_{\theta_1}^{\theta_2} \cos^2(3\theta) \, d\theta = \int_{\pi/6}^{5\pi/6} \cos^2(3\theta) \, d\theta$ To integrate $\cos^2(3\theta)$, we use the double angle identity: $\cos^2 x = \frac{1 + \cos(2x)}{2}$. So, $\cos^2(3\theta) = \frac{1 + \cos(2 \cdot 3\theta)}{2} = \frac{1 + \cos(6\theta)}{2}$. The integral becomes: $I = \int_{\pi/6}^{5\pi/6} \frac{1 + \cos(6\theta)}{2} \, d\theta$ $I = \frac{1}{2} \int_{\pi/6}^{5\pi/6} (1 + \cos(6\theta)) \, d\theta$ Now, we integrate term by term: $I = \frac{1}{2} \left[ \theta + \frac{\sin(6\theta)}{6} \right]_{\pi/6}^{5\pi/6}$ Now, we apply the limits of integration using the Fundamental Theorem of Calculus: $I = \frac{1}{2} \left[ \left( \frac{5\pi}{6} + \frac{\sin(6 \cdot \frac{5\pi}{6})}{6} \right) - \left( \frac{\pi}{6} + \frac{\sin(6 \cdot \frac{\pi}{6})}{6} \right) \right]$ $I = \frac{1}{2} \left[ \left( \frac{5\pi}{6} + \frac{\sin(5\pi)}{6} \right) - \left( \frac{\pi}{6} + \frac{\sin(\pi)}{6} \right) \right]$ We know that $\sin(5\pi) = 0$ and $\sin(\pi) = 0$. $I = \frac{1}{2} \left[ \left( \frac{5\pi}{6} + \frac{0}{6} \right) - \left( \frac{\pi}{6} + \frac{0}{6} \right) \right]$ $I = \frac{1}{2} \left[ \frac{5\pi}{6} - \frac{\pi}{6} \right]$ $I = \frac{1}{2} \left[ \frac{4\pi}{6} \right]$ $I = \frac{1}{2} \left[ \frac{2\pi}{3} \right]$ $I = \frac{\pi}{3}$

Common Mistakes & Tips

• Interpreting the Equation: Be careful with the notation. "$2\cot 2\theta$" could be ambiguous. In this case, the context and the solution imply it was $2\cot^2\theta$.
• Domain Restrictions: Always check if the solutions for $\theta$ fall within the given domain $(0, 2\pi) - \{\pi\}$. Also, ensure that terms in the original equation are well-defined for the found values of $\theta$.
• Trigonometric Identities for Integration: The identity $\cos^2 x = \frac{1 + \cos(2x)}{2}$ is crucial for integrating squared trigonometric functions.

Summary

We first transformed the given trigonometric equation into a quadratic equation in terms of $\sin \theta$. Solving this quadratic yielded possible values for $\sin \theta$. We then found the corresponding values of $\theta$ within the specified interval $(0, 2\pi) - \{\pi\}$. The smallest and largest of these values, $\theta_1$ and $\theta_2$, were identified. Finally, we evaluated the definite integral of $\cos^2(3\theta)$ from $\theta_1$ to $\theta_2$ by using a trigonometric identity to simplify the integrand and applying the fundamental theorem of calculus.

The final answer is \boxed{{\pi \over 3}}.