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

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


Алгебра | 6.4m просмотров
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+118 голосов

\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}

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