If $\tan (\theta)=\frac{3}{2}$, where $0 \leq \theta \leq 2 \pi$, then the largest positive value of $\theta$, to the nearest tenth, is rad.

Step 1 ：Given that \(\tan (\theta)=\frac{3}{2}\), where \(0 \leq \theta \leq 2 \pi\), we are asked to find the largest positive value of \(\theta\) in this range.

Step 2 ：The tangent function has a period of \(\pi\), which means it repeats its values every \(\pi\) radians. Therefore, if we find one value of \(\theta\) that satisfies the equation, we can add \(\pi\) to find another valid value of \(\theta\) in the given range.

Step 3 ：We start by finding the value of \(\theta\) in the first quadrant (i.e., \(0 \leq \theta \leq \frac{\pi}{2}\)) that satisfies the equation. This can be done by taking the arctangent of \(\frac{3}{2}\).

Step 4 ：Let's denote the first value of \(\theta\) as \(\theta_1\) and the second value as \(\theta_2\). We find that \(\theta_1 = 0.982793723247329\) and \(\theta_2 = 4.124386376837122\).

Step 5 ：The values of \(\theta\) obtained are in radians. The first value, \(\theta_1\), is in the first quadrant, and the second value, \(\theta_2\), is in the third quadrant.

Step 6 ：Since the question asks for the largest positive value of \(\theta\), the answer is \(\theta_2\). However, the question also asks for the answer to the nearest tenth. Therefore, we need to round \(\theta_2\) to the nearest tenth.

Step 7 ：We find that the rounded value of \(\theta_2\) is \(\theta_2_{rounded} = 4.1\).

Step 8 ：Final Answer: The largest positive value of \(\theta\), to the nearest tenth, is \(\boxed{4.1}\) radians.