To solve the equation
\[ \tan(2x) + \tan(3x) = \sec(3x), \]
we start by rewriting the equation in terms of sines and cosines. Recall that:
- \(\tan(x) = \frac{\sin(x)}{\cos(x)}\)
- \(\sec(x) = \frac{1}{\cos(x)}\)
Thus, the given equation can be expressed as:
\[ \frac{\sin(2x)}{\cos(2x)} + \frac{\sin(3x)}{\cos(3x)} = \frac{1}{\cos(3x)}. \]
Multiplying both sides by \(\cos(2x)\cos(3x)\):
\[ \sin(2x)\cos(3x) + \sin(3x)\cos(2x) = \cos(2x). \]
Next, apply the angle addition formula for sine:
\[ \sin(a + b) = \sin(a)\cos(b) + \cos(a)\sin(b). \]
Now we rewrite the left side:
\[ \sin(2x)\cos(3x) + \sin(3x)\cos(2x) = \sin(2x + 3x) = \sin(5x). \]
Thus, we have:
\[ \sin(5x) = \cos(2x). \]
To solve this equation, we can express \(\cos(2x)\) in terms of sine:
\[ \sin(5x) = \sin\left(\frac{\pi}{2} - 2x\right). \]
Using the sine equality, we have two cases to consider:
- \(5x = \frac{\pi}{2} - 2x + 2k\pi\) for some integer \(k\)
- \(5x = \pi - \left(\frac{\pi}{2} - 2x\right) + 2k\pi\) for some integer \(k\)
Case 1:
From \(5x = \frac{\pi}{2} - 2x + 2k\pi\):
Adding \(2x\) to both sides gives:
\[ 7x = \frac{\pi}{2} + 2k\pi \quad \Rightarrow \quad x = \frac{\pi}{14} + \frac{2k\pi}{7}. \]
Case 2:
From \(5x = \pi - \left(\frac{\pi}{2} - 2x\right) + 2k\pi\):
This simplifies to:
\[ 5x = \pi - \frac{\pi}{2} + 2x + 2k\pi \quad \Rightarrow \quad 5x = \frac{\pi}{2} + 2x + 2k\pi. \]
Subtracting \(2x\) from both sides gives:
\[ 3x = \frac{\pi}{2} + 2k\pi \quad \Rightarrow \quad x = \frac{\pi}{6} + \frac{2k\pi}{3}. \]
Finding the smallest positive solution
Now we need to evaluate the possible solutions for \(k = 0\):
- From Case 1: \(x = \frac{\pi}{14}\).
- From Case 2: \(x = \frac{\pi}{6}\).
We need to compare \(\frac{\pi}{14}\) and \(\frac{\pi}{6}\) to find the smallest positive root:
Calculating decimal approximations:
- \(\frac{\pi}{14} \approx 0.224\)
- \(\frac{\pi}{6} \approx 0.524\)
The smallest positive solution is thus:
\[ \boxed{\frac{\pi}{14}}. \]