The problem asks to find all values of $\theta$ in the interval $0 \leq \theta \leq \pi$ that satisfy the equation:

$\tan(2\theta) = \frac{4}{3}$

Step-by-step solution:

1. Solve for $2\theta$: We need to find the angle $\alpha$ such that:

   $\tan(\alpha) = \frac{4}{3}$

   Using the inverse tangent function (arctan or $\tan^{-1}$):

   $\alpha = \tan^{-1}\left(\frac{4}{3}\right)$

   From a calculator:

   $\alpha \approx 0.9273 \, \text{radians}$

2. Determine general solutions for $2\theta$: The tangent function has a period of $\pi$, meaning the general solution for $\alpha = 2\theta$ is:

   $2\theta = \alpha + n\pi \quad \text{where} \, n \in \mathbb{Z}$

   Therefore:

   $2\theta = 0.9273 + n\pi$

3. Solve for $\theta$: Now, divide both sides by 2 to solve for $\theta$:

   $\theta = \frac{0.9273 + n\pi}{2}$

4. Check for solutions within the interval $0 \leq \theta \leq \pi$:

   • For $n = 0$: $\theta = \frac{0.9273}{2} \approx 0.4637 \, \text{radians}$

   • For $n = 1$: $\theta = \frac{0.9273 + \pi}{2} \approx \frac{0.9273 + 3.1416}{2} = \frac{4.0689}{2} \approx 2.0344 \, \text{radians}$

     Since $2.0344$ is greater than $\pi$, it is outside the interval $0 \leq \theta \leq \pi$.

Final Answer:

The only solution in the given interval $0 \leq \theta \leq \pi$ is:

$\theta \approx 0.4637 \, \text{radians}$

Would you like any further details or clarifications?

Related Questions:

1. What is the period of the tangent function, and how does it affect the general solution?
2. How does the arctangent function return multiple values?
3. How can you verify the solution using a unit circle or calculator?
4. What would the solution look like if the interval was extended beyond $\pi$?
5. How can the same method be applied to different trigonometric functions?

Tip:

Always consider the periodicity of trigonometric functions when finding general solutions to equations involving them.