To solve the equation:

$\tan(3x) = \sqrt{3}$

we first recognize that we need to find angles for which the tangent equals $\sqrt{3}$.

Step 1: Identify the Reference Angle

The tangent of $\frac{\pi}{3}$ is $\sqrt{3}$. Thus:

$3x = \frac{\pi}{3} + n\pi$

where $n$ is an integer because the tangent function has a period of $\pi$.

Step 2: Solve for $x$

Now, we solve for $x$ by dividing both sides by 3:

$x = \frac{\pi}{9} + \frac{n\pi}{3}$

Step 3: Find All Solutions in the Interval $[0, \pi)$

We need to find values of $n$ such that $x$ remains within $[0, \pi)$.

1. For $n = 0$: $x = \frac{\pi}{9}$

2. For $n = 1$: $x = \frac{\pi}{9} + \frac{\pi}{3} = \frac{\pi}{9} + \frac{3\pi}{9} = \frac{4\pi}{9}$

3. For $n = 2$: $x = \frac{\pi}{9} + \frac{2\pi}{3} = \frac{\pi}{9} + \frac{6\pi}{9} = \frac{7\pi}{9}$

4. For $n = 3$: $x = \frac{\pi}{9} + \pi = \frac{\pi}{9} + \frac{9\pi}{9} = \frac{10\pi}{9}$ which is greater than $\pi$, so it is outside the interval $[0, \pi)$.

Final Answer

The exact solutions in the interval $[0, \pi)$ are:

$x = \frac{\pi}{9}, \frac{4\pi}{9}, \frac{7\pi}{9}$

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Would you like further details on any of these steps?

Related Questions

1. How would the solution change if the interval were $[0, 2\pi)$?
2. What are the general solutions for $\tan(3x) = -\sqrt{3}$?
3. How would we approach this problem if we had $\tan(3x) = 1$ instead?
4. What is the period of $\tan(3x)$, and how does it affect the solution set?
5. Can this method be applied to solve $\tan(kx) = c$ for any constant $k$ and $c$?

Tip

When working with trigonometric equations, always consider the periodicity of the trigonometric function to find all solutions within a specified interval.