Given that tan(x − π/3) = 1, solve x for x ∈(0°, 180°)

$\tan \left(x - \dfrac{\pi}{3} \right) = 1$

Replacement: $x - \dfrac{\pi}{3} = t$

$\tan (t) = 1$

$t = \tan^{-1}(1) + \pi k, \ k \in \mathbb{Z}$

$t = \dfrac{\pi}{4} + \pi k, \ k \in \mathbb{Z}$

Reverse replacement:

$x - \dfrac{\pi}{3} = \dfrac{\pi}{4} + \pi k, \ k \in \mathbb{Z}$

$x - \dfrac{\pi}{3} + \dfrac{\pi}{3} = \dfrac{\pi}{4} + \dfrac{\pi}{3} + \pi k, \ k \in \mathbb{Z}$

$x = \dfrac{7 \pi}{12} + \pi k, \ k \in \mathbb{Z}$

Since $x \in (0^{\circ}; \ 180^{\circ})$ i. e. $x \in (0; \ \pi)$ then:

$0 < \dfrac{7 \pi}{12} + \pi k < \pi, \ k \in \mathbb{Z}$

$0 - \dfrac{7 \pi}{12} < \dfrac{7 \pi}{12} - \dfrac{7 \pi}{12} + \pi k < \pi - \dfrac{7 \pi}{12}, \ k \in \mathbb{Z}$

$-\dfrac{7 \pi}{12} < \pi k < \dfrac{5 \pi}{12}, \ k \in \mathbb{Z}$

$-\dfrac{7 \pi}{12} \cdot \dfrac{1}{\pi} < \pi k \cdot \dfrac{1}{\pi} < \dfrac{5 \pi}{12} \cdot \dfrac{1}{\pi}, \ k \in \mathbb{Z}$

$-\dfrac{7}{12} < k < \dfrac{5}{12}, \ k \in \mathbb{Z}$

Thus $k = 0$

Therefore, $x = \dfrac{7 \pi}{12} + \pi \cdot 0 = \dfrac{7 \pi}{12}$

Answer: $x = \dfrac{7 \pi}{12}$