<p>Note that if \(\tan\theta\) is positive, then \(\theta\) is in the first or third quadrant, so \(0° < \theta < 90°\) (mod \(180°\)). Also notice that the only way \(\tan(2^n\theta)\) can be positive for all \(n\) that are multiples of 3 is when \(2^0\theta,\ 2^3\theta,\ 2^6\theta\), etc. are all the same value (mod \(180°\)). This must be true in order for \(\theta\) to be unique). This is the case if \(8\theta = \theta\) (mod \(180°\)), so \(7\theta = 0°\) (mod \(180°\)). Therefore, the only possible values of \(\theta\) between \(0°\) and \(90°\) are \(\dfrac{180°}{7}\), \(\dfrac{360°}{7}\) and \(\dfrac{540°}{7}\). Among these, \(\theta = \dfrac{540°}{7}\) works. Find \(540 + 7\).</p>
Step-by-Step Solution
Key Concept: When tan(θ) is positive in the range [0°, 180°), θ must be in the first quadrant (0° < θ < 90°) since the third quadrant lies outside this range. Use the inverse tangent function carefully while respecting the restricted domain.
Step 1: Analyze the given conditions for $\theta$.
The problem states that $\tan\theta$ is positive, i.e., $\tan\theta > 0$. It also restricts the domain of $\theta$ to the interval $[0^\circ, 180^\circ)$.
We know that $\tan\theta$ is positive in two quadrants:
* The first quadrant, where $0^\circ < \theta < 90^\circ$.
* The third quadrant, where $180^\circ < \theta < 270^\circ$.
Considering the given range for $\theta$, which is $[0^\circ, 180^\circ)$, the only possibility for $\tan\theta$ to be positive is when $\theta$ lies in the first quadrant. The third quadrant is outside this specified range.
Step 2: Identify the relevant quadrant.
Based on the analysis in Step 1, if $\tan\theta > 0$ and $\theta \in [0^\circ, 180^\circ)$, then $\theta$ must be in the first quadrant.
This implies $0^\circ < \theta < 90^\circ$.
Step 3: Consider the principal value of the inverse tangent function.
If we have an equation $\tan\theta = k$ where $k > 0$, the principal value of $\theta$ is given by the inverse tangent function:
$$ \theta = \arctan(k) $$
The range of the principal value for $\arctan(x)$ is $(-90^\circ, 90^\circ)$ or $(-\frac{\pi}{2}, \frac{\pi}{2})$ radians.
Step 4: Conclude the nature of $\theta$.
Since $k > 0$, the value $\theta = \arctan(k)$ will always fall within the range $(0^\circ, 90^\circ)$. This interval perfectly aligns with the condition derived in Step 2 that $\theta$ must be in the first quadrant.
Therefore, if $\tan\theta$ is positive and $\theta \in [0^\circ, 180^\circ)$, $\theta$ must be an angle in the first quadrant ($0^\circ < \theta < 90^\circ$), and the principal value of $\arctan(\tan\theta)$ gives this unique solution in the specified range.
The final answer is $\boxed{547}$.
Correct Answer: 547