We are given two conditions involving angles $\alpha$ and $\beta$ within the interval $0^\circ < x < 180^\circ$:

1. $\tan \alpha = 2 \tan \beta$
2. $\tan(\alpha + \beta) = 3$

We need to find the possible values of $\alpha$ and $\beta$.

Step 1: Use the tangent addition formula

The tangent of a sum is given by: $\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$ Substitute the known value $\tan(\alpha + \beta) = 3$: $3 = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$ This equation will allow us to relate $\tan \alpha$ and $\tan \beta$.

Step 2: Use $\tan \alpha = 2 \tan \beta$

From the first equation, $\tan \alpha = 2 \tan \beta$. Let $\tan \beta = t$. Then, $\tan \alpha = 2t$.

Substitute these into the sum formula: $3 = \frac{2t + t}{1 - 2t \cdot t}$ This simplifies to: $3 = \frac{3t}{1 - 2t^2}$

Step 3: Solve for $t$

Now solve the equation: $3(1 - 2t^2) = 3t$ Expanding both sides: $3 - 6t^2 = 3t$ Rearrange it into a standard quadratic form: $6t^2 + 3t - 3 = 0$ Simplify by dividing by 3: $2t^2 + t - 1 = 0$

Step 4: Solve the quadratic equation

We now solve $2t^2 + t - 1 = 0$ using the quadratic formula: $t = \frac{-1 \pm \sqrt{1^2 - 4(2)(-1)}}{2(2)}$ $t = \frac{-1 \pm \sqrt{1 + 8}}{4}$ $t = \frac{-1 \pm \sqrt{9}}{4}$ $t = \frac{-1 \pm 3}{4}$ Thus, $t = \frac{2}{4} = \frac{1}{2}$ or $t = \frac{-4}{4} = -1$.

Step 5: Find $\tan \alpha$ and $\tan \beta$

We have two cases for $t$:

1. If $t = \frac{1}{2}$, then $\tan \beta = \frac{1}{2}$ and $\tan \alpha = 2 \times \frac{1}{2} = 1$.
2. If $t = -1$, then $\tan \beta = -1$ and $\tan \alpha = 2 \times (-1) = -2$.

Step 6: Find $\alpha$ and $\beta$

Now, we find the angles corresponding to these tangent values:

• For $\tan \beta = \frac{1}{2}$ and $\tan \alpha = 1$: $\beta = \tan^{-1}\left(\frac{1}{2}\right) \approx 26.57^\circ$ $\alpha = \tan^{-1}(1) = 45^\circ$

• For $\tan \beta = -1$ and $\tan \alpha = -2$: $\beta = \tan^{-1}(-1) = 135^\circ \quad (\text{since} 180^\circ > \beta > 90^\circ)$ $\alpha = \tan^{-1}(-2) \approx 116.57^\circ$

Final Solutions:

The possible values of $\alpha$ and $\beta$ are:

1. $\alpha = 45^\circ$, $\beta = 26.57^\circ$
2. $\alpha = 116.57^\circ$, $\beta = 135^\circ$

Would you like further clarification or more details on any step?

Here are 5 related questions for further exploration:

1. What is the general solution for $\alpha$ and $\beta$ in terms of periodicity for tangent?
2. How would the values change if the interval was extended beyond $0^\circ < x < 180^\circ$?
3. How can we use trigonometric identities to verify the solutions obtained?
4. Can you derive the tangent sum formula from the basic sine and cosine identities?
5. What would happen if the given equations involved sine or cosine instead of tangent?

Tip: Remember that the tangent function repeats every 180°, so always consider possible periodic solutions in such trigonometric problems.