Solve $\sqrt{3} \tan^2 \theta + 3 \tan \theta > 0$

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Hey there, math enthusiasts! Welcome back to Plastik Magazine, where we dive deep into the coolest math problems. Today, we're tackling a trigonometric inequality that might look a little intimidating at first glance, but trust me, guys, with a little bit of algebraic finesse and trigonometric know-how, we'll conquer it. The inequality we're looking at is $\sqrt{3} \tan^2 \theta + 3 \tan \theta > 0$. This problem is all about understanding how the tangent function behaves and how to solve inequalities involving it. We'll break it down step-by-step, making sure to cover all the bases so you can feel confident solving similar problems on your own. Get ready to flex those math muscles!

Understanding the Inequality and Initial Steps

Alright, let's get straight to it. Our main goal is to solve the inequality $\sqrt{3} \tan^2 \theta + 3 \tan \theta > 0$. The first thing we should notice is that this inequality has a common factor, which is $\tan \theta$. Factoring this out will simplify the expression significantly and make it much easier to analyze. So, let's factor out $\tan \theta$ from both terms:

$\qquad \tan \theta (\sqrt{3} \tan \theta + 3) > 0$

Now, this inequality holds true if either both factors are positive or both factors are negative. This gives us two distinct cases to consider. This is a crucial step, as it transforms a single complex inequality into two simpler inequalities that we can solve independently. Remember, when you're faced with an inequality like this, always look for common factors or ways to simplify the expression. It's like finding a hidden shortcut on a tricky path!

Case 1: Both factors are positive.

This means we need both $\tan \theta > 0$ AND $\sqrt{3} \tan \theta + 3 > 0$. Let's break down the second part of this case: $\sqrt{3} \tan \theta + 3 > 0$. To solve for $\tan \theta$, we subtract 3 from both sides:

$\qquad \sqrt{3} \tan \theta > -3$

Now, divide by $\sqrt{3}$:

$\qquad \tan \theta > \frac{-3}{\sqrt{3}}$

To simplify the right side, we can multiply the numerator and denominator by $\sqrt{3}$:

$\qquad \tan \theta > \frac{-3\sqrt{3}}{3}$

$\qquad \tan \theta > -\sqrt{3}$

So, for Case 1, we need to satisfy two conditions simultaneously: $\tan \theta > 0$ AND $\tan \theta > -\sqrt{3}$. When we consider both of these, the more restrictive condition is $\tan \theta > 0$. This is because any value of $\tan \theta$ that is greater than 0 is automatically greater than -\sqrt{3}. So, Case 1 simplifies to just $\tan \theta > 0$.

Case 2: Both factors are negative.

This means we need both $\tan \theta < 0$ AND $\sqrt{3} \tan \theta + 3 < 0$. Let's solve the second inequality in this case: $\sqrt{3} \tan \theta + 3 < 0$. Subtracting 3 from both sides gives:

$\qquad \sqrt{3} \tan \theta < -3$

Dividing by $\sqrt{3}$:

$\qquad \tan \theta < \frac{-3}{\sqrt{3}}$

And as we simplified before, this becomes:

$\qquad \tan \theta < -\sqrt{3}$

So, for Case 2, we need to satisfy two conditions simultaneously: $\tan \theta < 0$ AND $\tan \theta < -\sqrt{3}$. When we consider both of these, the more restrictive condition is $\tan \theta < -\sqrt{3}$. This is because any value of $\tan \theta$ that is less than -\sqrt{3} is automatically less than 0. So, Case 2 simplifies to $\tan \theta < -\sqrt{3}$.

Now we have successfully broken down our original inequality into two simpler conditions: $\tan \theta > 0$ or $\tan \theta < -\sqrt{3}$. The next step is to figure out for which values of $\theta$ these conditions are met. This involves understanding the unit circle and the behavior of the tangent function across different quadrants. Keep your eyes peeled for the next section where we'll map these conditions onto the unit circle!

Analyzing the Tangent Function and its Behavior

To find the values of $\theta$ that satisfy our conditions, $\tan \theta > 0$ and $\tan \theta < -\sqrt{3}$, we need to analyze the behavior of the tangent function. Remember, $\tan \theta = \frac{\sin \theta}{\cos \theta}$. The tangent function is positive in the quadrants where both $\sin \theta$ and $\cos \theta$ have the same sign, and negative where they have opposite signs. Specifically:

• Quadrant I (0 to $\frac{\pi}{2}$): $\sin \theta > 0$, $\cos \theta > 0$. Therefore, $\tan \theta > 0$.
• Quadrant II ($\frac{\pi}{2}$ to $\pi$): $\sin \theta > 0$, $\cos \theta < 0$. Therefore, $\tan \theta < 0$.
• Quadrant III ($\pi$ to $\frac{3\pi}{2}$): $\sin \theta < 0$, $\cos \theta < 0$. Therefore, $\tan \theta > 0$.
• Quadrant IV ($\frac{3\pi}{2}$ to $2\pi$): $\sin \theta < 0$, $\cos \theta > 0$. Therefore, $\tan \theta < 0$.

Also, remember that the tangent function has vertical asymptotes at $\theta = \frac{\pi}{2} + n\pi$, where $n$ is an integer. This is because $\cos \theta = 0$ at these points, making $\tan \theta$ undefined.

Let's consider our first condition: $\tan \theta > 0$. Based on our quadrant analysis, this occurs in Quadrant I and Quadrant III. Within a standard interval of $0$ to $2\pi$ (excluding the asymptotes), this means:

• For Quadrant I: $0 < \theta < \frac{\pi}{2}$
• For Quadrant III: $\pi < \theta < \frac{3\pi}{2}$

Now, let's consider our second condition: $\tan \theta < -\sqrt{3}$. We need to find the angles where the tangent value is less than -\sqrt{3}. We know that $\tan \theta = -\sqrt{3}$ at specific angles. The reference angle for which $\tan \alpha = \sqrt{3}$ is $\alpha = \frac{\pi}{3}$. Since the tangent is negative in Quadrant II and Quadrant IV, the angles where $\tan \theta = -\sqrt{3}$ within the interval $[0, 2\pi)$ are:

• In Quadrant II: $\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$
• In Quadrant IV: $\theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$

Now we need $\tan \theta < -\sqrt{3}$.

• In Quadrant II, the tangent decreases as $\theta$ increases towards $\frac{\pi}{2}$. So, for $\tan \theta < -\sqrt{3}$, we need $\frac{\pi}{2} < \theta < \frac{2\pi}{3}$. (Remember $\tan \theta$ is undefined at $\frac{\pi}{2}$ and approaches negative infinity as $\theta$ approaches $\frac{\pi}{2}$ from the right).
• In Quadrant IV, the tangent increases as $\theta$ increases towards $\frac{3\pi}{2}$. So, for $\tan \theta < -\sqrt{3}$, we need $\frac{5\pi}{3} < \theta < \frac{3\pi}{2}$. (Remember $\tan \theta$ is undefined at $\frac{3\pi}{2}$ and approaches negative infinity as $\theta$ approaches $\frac{3\pi}{2}$ from the right).

So, combining these, the condition $\tan \theta < -\sqrt{3}$ is satisfied for $\frac{\pi}{2} < \theta < \frac{2\pi}{3}$ and $\frac{5\pi}{3} < \theta < \frac{3\pi}{2}$.

We have now identified the intervals for both conditions: $\tan \theta > 0$ and $\tan \theta < -\sqrt{3}$. The solution to our original inequality is the union of the intervals where either of these conditions is met. This means we are looking for angles $\theta$ that fall into the intervals for $\tan \theta > 0$ OR the intervals for $\tan \theta < -\sqrt{3}$. This is a key point: we need to combine the solution sets of the two cases we analyzed earlier. It's like saying, "We're happy if condition A is met, OR if condition B is met." Let's visualize this on the unit circle or a number line to get the final answer!

Visualizing the Solution on the Unit Circle

Let's bring it all together by visualizing our solutions. We need to find all values of $\theta$ for which $\tan \theta > 0$ OR $\tan \theta < -\sqrt{3}$. We'll consider the principal interval of $0 \le \theta < 2\pi$, and then we can generalize it.

Condition 1: $\tan \theta > 0$

This occurs when $\theta$ is in Quadrant I or Quadrant III. Specifically:

• Quadrant I: $0 < \theta < \frac{\pi}{2}$
• Quadrant III: $\pi < \theta < \frac{3\pi}{2}$

Condition 2: $\tan \theta < -\sqrt{3}$

We found that $\tan \theta = -\sqrt{3}$ at $\theta = \frac{2\pi}{3}$ and $\theta = \frac{5\pi}{3}$.

• For $\tan \theta < -\sqrt{3}$ in Quadrant II, we need angles between $\frac{\pi}{2}$ (where tangent is undefined and goes to $-\infty$) and $\frac{2\pi}{3}$. So, $\frac{\pi}{2} < \theta < \frac{2\pi}{3}$.
• For $\tan \theta < -\sqrt{3}$ in Quadrant IV, we need angles between $\frac{5\pi}{3}$ and $\frac{3\pi}{2}$ (where tangent is undefined and goes to $-\infty$ as it approaches from the left). So, $\frac{5\pi}{3} < \theta < \frac{3\pi}{2}$.

Now, we take the union of all these intervals because our original inequality is satisfied if either condition is met.

The intervals where $\tan \theta > 0$ are $(0, \frac{\pi}{2})$ and $(\pi, \frac{3\pi}{2})$. The intervals where $\tan \theta < -\sqrt{3}$ are $(\frac{\pi}{2}, \frac{2\pi}{3})$ and $(\frac{5\pi}{3}, \frac{3\pi}{2})$.

So, the solution set for $\theta$ in the interval $[0, 2\pi)$ is:

• From Condition 1: $(0, \frac{\pi}{2}) \cup (\pi, \frac{3\pi}{2})$
• From Condition 2: $(\frac{\pi}{2}, \frac{2\pi}{3}) \cup (\frac{5\pi}{3}, \frac{3\pi}{2})$

Combining these, we get:

$\qquad (0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \frac{2\pi}{3}) \cup (\pi, \frac{3\pi}{2}) \cup (\frac{5\pi}{3}, \frac{3\pi}{2})$

Wait a minute! Let's look closer at the interval $(0, \frac{\pi}{2})$ and $(\frac{\pi}{2}, \frac{2\pi}{3})$. Notice that $\frac{\pi}{2}$ is an asymptote for the tangent function, so $\tan \theta$ is undefined there. However, the intervals are adjacent and include all values where $\tan \theta > 0$ or $\tan \theta < -\sqrt{3}$.

Let's rethink the combination of intervals. We are looking for $\theta$ such that $\tan \theta > 0$ OR $\tan \theta < -\sqrt{3}$.

The intervals where $\tan \theta > 0$ are $n\pi < \theta < \frac{\pi}{2} + n\pi$ for any integer $n$.

The intervals where $\tan \theta < -\sqrt{3}$ are $\frac{\pi}{2} + n\pi < \theta < \frac{2\pi}{3} + n\pi$ for any integer $n$ (this accounts for the quadrants where tangent is negative and values below $-\sqrt{3}$).

Let's look at a single period, say $0 \le \theta < 2\pi$:

• For $n=0$: $\tan \theta > 0$ in $(0, \frac{\pi}{2})$ and $(\pi, \frac{3\pi}{2})$. $\tan \theta < -\sqrt{3}$ in $(\frac{\pi}{2}, \frac{2\pi}{3})$ and $(\frac{5\pi}{3}, \frac{3\pi}{2})$.

Wait, this combination is not quite right. Let's go back to the conditions: $\tan \theta > 0$ OR $\tan \theta < -\sqrt{3}$.

Angles where $\tan \theta = 0$: $\theta = n\pi$ Angles where $\tan \theta = -\sqrt{3}$: $\theta = \frac{2\pi}{3} + n\pi$ Angles where $\tan \theta$ is undefined (vertical asymptotes): $\theta = \frac{\pi}{2} + n\pi$

We need to examine the intervals defined by these critical points.

Consider the interval $(0, 2\pi)$. The critical points are $0, \frac{\pi}{2}, \frac{2\pi}{3}, \pi, \frac{3\pi}{2}, \frac{5\pi}{3}, 2\pi$.

Let's test values in the intervals:

• $(0, \frac{\pi}{2})$: $\tan \theta > 0$. This interval works.
• $(\frac{\pi}{2}, \frac{2\pi}{3})$: $\tan \theta < -\sqrt{3}$. This interval works.
• $(\frac{2\pi}{3}, \pi)$: $\tan \theta$ is between $-\sqrt{3}$ and $0$. This interval does not work.
• $(\pi, \frac{3\pi}{2})$: $\tan \theta > 0$. This interval works.
• $(\frac{3\pi}{2}, \frac{5\pi}{3})$: $\tan \theta$ is between $0$ and $-\sqrt{3}$ (as $\theta$ approaches $\frac{3\pi}{2}$ from the right, $\tan \theta$ goes to $-\infty$, and at $\frac{5\pi}{3}$, $\tan \theta = -\sqrt{3}$). So, in this interval, $\tan \theta < -\sqrt{3}$. This interval works.
• $(\frac{5 ext{\pi}}{3}, 2\pi)$: $\tan \theta$ is between $-\sqrt{3}$ and $0$. This interval does not work.

Therefore, within the interval $[0, 2\pi)$, the solution is:

$\qquad (0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \frac{2\pi}{3}) \cup (\pi, \frac{3\pi}{2}) \cup (\frac{5\pi}{3}, \frac{3\pi}{2})$

Looking closer at the intervals from Case 1 and Case 2:

Case 1 solution: $\tan \theta > 0$, which is $n\pi < \theta < \frac{\pi}{2} + n\pi$. For $n=0$, $(0, \frac{\pi}{2})$ and $(\pi, \frac{3\pi}{2})$. For $n=1$, $(\pi, \frac{3\pi}{2})$ and $(\frac{3\pi}{2}, 2\pi)$.

Case 2 solution: $\tan \theta < -\sqrt{3}$, which is $\frac{\pi}{2} + n\pi < \theta < \frac{2\pi}{3} + n\pi$. For $n=0$, $(\frac{\pi}{2}, \frac{2\pi}{3})$. For $n=1$, $(\frac{3\pi}{2}, \frac{5\pi}{3})$.

Combining all intervals from $0$ to $2\pi$ where $\tan \theta > 0$ or $\tan \theta < -\sqrt{3}$:

• We need $\tan \theta > 0$. This covers $(0, \frac{\pi}{2})$ and $(\pi, \frac{3\pi}{2})$.
• We need $\tan \theta < -\sqrt{3}$. This covers $(\frac{\pi}{2}, \frac{2\pi}{3})$ and $(\frac{3\pi}{2}, \frac{5\pi}{3})$.

Ah, I see the error in my interval testing. The tangent function has a period of $\pi$. Let's stick to analyzing the behavior within a single period and then generalize.

Let's re-evaluate the intervals based on $\tan \theta > 0$ OR $\tan \theta < -\sqrt{3}$.

The angles where $\tan \theta = 0$ are $0, \pi, 2\pi, ...$

The angles where $\tan \theta = -\sqrt{3}$ are $\frac{2\pi}{3}, \frac{5\pi}{3}, ...$

The vertical asymptotes are $\frac{\pi}{2}, \frac{3\pi}{2}, ...$

Consider the interval $(0, \pi)$.

• $(0, \frac{\pi}{2})$: $\tan \theta > 0$. This interval works.
• $(\frac{\pi}{2}, \frac{2\pi}{3})$: $\tan \theta < -\sqrt{3}$. This interval works.
• $(\frac{2\pi}{3}, \pi)$: $-\sqrt{3} < \tan \theta < 0$. This interval does not work.

So, for the interval $(0, \pi)$, the solution is $(0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \frac{2\pi}{3})$.

Since the tangent function has a period of $\pi$, we can add $n\pi$ to these intervals to get the general solution.

General Solution:

$\qquad n\pi < \theta < \frac{\pi}{2} + n\pi$ (This is where $\tan \theta > 0$)

OR

$\qquad \frac{\pi}{2} + n\pi < \theta < \frac{2\pi}{3} + n\pi$ (This is where $\tan \theta < -\sqrt{3}$)

Let's check this again. The condition is $\tan \theta (\sqrt{3} \tan \theta + 3) > 0$. This means $(\tan \theta > 0 ext{ and } \sqrt{3} \tan \theta + 3 > 0)$ OR $(\tan \theta < 0 ext{ and } \sqrt{3} \tan \theta + 3 < 0)$.

We found that $\sqrt{3} \tan \theta + 3 > 0$ implies $\tan \theta > -\sqrt{3}$. We found that $\sqrt{3} \tan \theta + 3 < 0$ implies $\tan \theta < -\sqrt{3}$.

So, the conditions become:

1. $(\tan \theta > 0 ext{ and } \tan \theta > -\sqrt{3})$ which simplifies to $\tan \theta > 0$.
2. $(\tan \theta < 0 ext{ and } \tan \theta < -\sqrt{3})$ which simplifies to $\tan \theta < -\sqrt{3}$.

This brings us back to: $\tan \theta > 0$ OR $\tan \theta < -\sqrt{3}$.

Let's use the unit circle and the properties of tangent. Tangent is positive in Quadrant I and III. Tangent is negative in Quadrant II and IV.

Intervals for $\tan \theta > 0$:

• Quadrant I: $(0, \frac{\pi}{2})$
• Quadrant III: $(\pi, \frac{3\pi}{2})$

Intervals for $\tan \theta < -\sqrt{3}$:

We know $\tan \theta = -\sqrt{3}$ at $\theta = \frac{2\pi}{3}$ and $\theta = \frac{5\pi}{3}$.

• In Quadrant II, tangent goes from $0$ to $-\infty$ as $\theta$ goes from $\pi$ to $\frac{\pi}{2}$. So, $\tan \theta < -\sqrt{3}$ when $\frac{\pi}{2} < \theta < \frac{2\pi}{3}$.
• In Quadrant IV, tangent goes from $-\infty$ to $0$ as $\theta$ goes from $\frac{3\pi}{2}$ to $2\pi$. So, $\tan \theta < -\sqrt{3}$ when $\frac{3\pi}{2} < \theta < \frac{5\pi}{3}$.

Now, we take the union of all these intervals where $\tan \theta > 0$ OR $\tan \theta < -\sqrt{3}$.

Looking at $0 \le \theta < 2\pi$:

• $(0, \frac{\pi}{2})$ (from $\tan \theta > 0$)
• $(\frac{\pi}{2}, \frac{2\pi}{3})$ (from $\tan \theta < -\sqrt{3}$)
• $(\pi, \frac{3\pi}{2})$ (from $\tan \theta > 0$)
• $(\frac{3\pi}{2}, \frac{5\pi}{3})$ (from $\tan \theta < -\sqrt{3}$)

So, the solution in $[0, 2\pi)$ is $(0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \frac{2\pi}{3}) \cup (\pi, \frac{3\pi}{2}) \cup (\frac{3\pi}{2}, \frac{5\pi}{3})$.

This looks correct! We've covered all the regions where the inequality holds.

Generalizing the Solution

Now that we've found the solution within the interval $[0, 2\pi)$, we can generalize it for all real numbers $\theta$. The tangent function has a period of $\pi$. This means that the pattern of values repeats every $\pi$ radians. However, our solution intervals are not uniformly distributed with respect to $\pi$. We have intervals in Quadrant I and II, and then in Quadrant III and IV.

Let's consider the general form of our intervals:

1. $n\pi < \theta < \frac{\pi}{2} + n\pi$ (This covers Quadrant I and III where $\tan \theta > 0$)
2. $\frac{\pi}{2} + n\pi < \theta < \frac{2\pi}{3} + n\pi$ (This covers Quadrant II where $\tan \theta < -\sqrt{3}$)
3. $\frac{3\pi}{2} + n\pi < \theta < \frac{5\pi}{3} + n\pi$ (This covers Quadrant IV where $\tan \theta < -\sqrt{3}$)

This way of expressing it seems a bit redundant and potentially confusing due to the $n\pi$ shifts.

Let's go back to the simplified conditions: $\tan \theta > 0$ or $\tan \theta < -\sqrt{3}$.

For $\tan \theta > 0$: This occurs in Quadrant I and III. The general solution is $n\pi < \theta < \frac{\pi}{2} + n\pi$, where $n$ is an integer.

For $\tan \theta < -\sqrt{3}$: This occurs in Quadrant II and IV. The angles where $\tan \theta = -\sqrt{3}$ are $\theta = \frac{2\pi}{3} + k\pi$. We need values less than $-\sqrt{3}$. In Quadrant II (where $\pi/2 < \theta < \pi$), we need $\frac{\pi}{2} < \theta < \frac{2\pi}{3}$. In Quadrant IV (where $3\pi/2 < \theta < 2\pi$), we need $\frac{3\pi}{2} < \theta < \frac{5\pi}{3}$.

If we consider the general form for $\tan \theta < -\sqrt{3}$, we need to be careful with the period. The interval $(\frac{\pi}{2}, \frac{2\pi}{3})$ has a length of $\frac{\pi}{6}$. The interval $(\frac{3\pi}{2}, \frac{5\pi}{3})$ also has a length of $\frac{\pi}{6}$.

A more direct way to express the solution for $\tan \theta < -\sqrt{3}$ is to consider the reference angle. The primary angle where $\tan \theta = -\sqrt{3}$ is $\frac{2\pi}{3}$. The tangent function is decreasing in the interval $(\frac{\pi}{2}, \frac{3\pi}{2})$.

Let's use the intervals we found in $[0, 2\pi)$: $(0, \frac{\pi}{2})$, $(\frac{\pi}{2}, \frac{2\pi}{3})$, $(\pi, \frac{3\pi}{2})$, $(\frac{3\pi}{2}, \frac{5\pi}{3})$.

Notice that the structure of the solution within $(0, \pi)$ is $(0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \frac{2\pi}{3})$. The structure of the solution within $(\pi, 2\pi)$ is $(\pi, \frac{3\pi}{2}) \cup (\frac{3\pi}{2}, \frac{5\pi}{3})$.

This means that for every interval of length $\pi$, we have the same pattern:

$\qquad n\pi < \theta < \frac{\pi}{2} + n\pi$ (This covers the first part of the interval where $\tan \theta > 0$)

and

$\qquad \frac{\pi}{2} + n\pi < \theta < \frac{2\pi}{3} + n\pi$ (This covers the second part where $\tan \theta < -\sqrt{3}$)

However, this still feels a bit off. Let's re-examine the regions. We need $\tan \theta > 0$ OR $\tan \theta < -\sqrt{3}$.

Consider the graph of $y = \tan \theta$. We are looking for regions where the graph is above the x-axis OR below the line $y = -\sqrt{3}$.

Regions where $\tan \theta > 0$:

• $n\pi < \theta < \frac{\pi}{2} + n\pi$

Regions where $\tan \theta < -\sqrt{3}$: Let $\alpha$ be the angle in $(0, \pi/2)$ such that $\tan \alpha = \sqrt{3}$. So, $\alpha = \frac{\pi}{3}$. Tangent is negative in Quadrant II and IV.

• In Quadrant II, $\tan \theta = -\sqrt{3}$ at $\theta = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$. For $\tan \theta < -\sqrt{3}$, we need $\frac{\pi}{2} < \theta < \frac{2\pi}{3}$.
• In Quadrant IV, $\tan \theta = -\sqrt{3}$ at $\theta = 2\pi - \frac{\pi}{3} = \frac{5\pi}{3}$. For $\tan \theta < -\sqrt{3}$, we need $\frac{3\pi}{2} < \theta < \frac{5\pi}{3}$.

Now, let's combine these general solutions. We are looking for the union of:

$\qquad (n\pi, \frac{\pi}{2} + n\pi)$ AND $(\frac{\pi}{2} + n\pi, \frac{2\pi}{3} + n\pi)$ AND $(\frac{3\pi}{2} + n\pi, \frac{5\pi}{3} + n\pi)$

This still looks a bit messy. Let's organize by the behavior of tangent.

Tangent is positive in $(n\pi, \frac{\pi}{2} + n\pi)$. Tangent is negative in $(\frac{\pi}{2} + n\pi, (n+1)\pi)$.

We need $\tan \theta > 0$ OR $\tan \theta < -\sqrt{3}$.

Case A: $\tan \theta > 0$ This occurs for $n\pi < \theta < \frac{\pi}{2} + n\pi$ for any integer $n$.

Case B: $\tan \theta < -\sqrt{3}$ This occurs in the intervals where tangent is negative and the value is less than $-\sqrt{3}$. Consider an interval $(\frac{\pi}{2} + n\pi, (n+1)\pi)$. Here, $\tan \theta$ is negative. We know $\tan \theta = -\sqrt{3}$ at $\theta = \frac{2\pi}{3} + k\pi$. We need $\tan \theta < -\sqrt{3}$. So, in the interval $(\frac{\pi}{2} + n\pi, (n+1)\pi)$, the condition $\tan \theta < -\sqrt{3}$ is met when $\frac{\pi}{2} + n\pi < \theta < \frac{2\pi}{3} + n\pi$.

So, the complete solution is the union of the intervals from Case A and Case B:

$\qquad n\pi < \theta < \frac{\pi}{2} + n\pi$ (for any integer $n$)

OR

$\qquad \frac{\pi}{2} + n\pi < \theta < \frac{2\pi}{3} + n\pi$ (for any integer $n$)

Let's verify this. If $n=0$: $0 < \theta < \frac{\pi}{2}$ OR $\frac{\pi}{2} < \theta < \frac{2\pi}{3}$. This matches our earlier findings for the interval (0, rac{2 ext{\pi}}{3}) excluding $\frac{\pi}{2}$.

If $n=1$: $\pi < \theta < \frac{3\pi}{2}$ OR $\frac{3\pi}{2} < \theta < \frac{5\pi}{3}$. This matches our earlier findings for the interval $(\pi, \frac{5\text{\pi}}{3})$ excluding $\frac{3\pi}{2}$.

This looks like the correct general solution. We've combined the regions where the inequality $\sqrt{3} \tan^2 \theta + 3 \tan \theta > 0$ holds true.

Final Answer and Key Takeaways

To sum up, we started with the inequality $\sqrt{3} \tan^2 \theta + 3 \tan \theta > 0$. By factoring out $\tan \theta$, we got $\tan \theta (\sqrt{3} \tan \theta + 3) > 0$. This led us to two main conditions: $\tan \theta > 0$ or $\tan \theta < -\sqrt{3}$.

We analyzed the behavior of the tangent function and found that:

• $\tan \theta > 0$ in the intervals $n\pi < \theta < \frac{\pi}{2} + n\pi$ for any integer $n$.
• $\tan \theta < -\sqrt{3}$ in the intervals $\frac{\pi}{2} + n\pi < \theta < \frac{2\pi}{3} + n\pi$ for any integer $n$.

Combining these, the general solution to the inequality is the union of these two sets of intervals:

$\qquad \boxed{\bigcup_{n \in \mathbb{Z}} \left( n\pi < \theta < \frac{\pi}{2} + n\pi \quad \text{or} \quad \frac{\pi}{2} + n\pi < \theta < \frac{2\pi}{3} + n\pi \right)}$

Some key takeaways from this problem, guys:

1. Factorization is your friend: Always look for common factors in inequalities. It often simplifies the problem immensely.
2. Case analysis: When dealing with products or quotients in inequalities, breaking it down into cases (e.g., both positive, both negative) is a powerful technique.
3. Understand function behavior: Knowing where a trigonometric function is positive, negative, increasing, or decreasing is crucial for solving inequalities.
4. Unit Circle / Graph: Visualizing the solutions on the unit circle or by sketching the graph of the function can help confirm your interval analysis.
5. Generalization: Remember to use the periodicity of trigonometric functions to express the general solution.

Solving trigonometric inequalities requires a solid understanding of trigonometric identities, function behavior, and algebraic manipulation. Keep practicing, and you'll master these challenges in no time! If you found this breakdown helpful, give it a share and let us know what other math mysteries you'd like us to unravel. Until next time, happy solving!