Explanation:
If the question is like this,
![\tan( (\pi)/(3)x - 3 ) = 0](https://img.qammunity.org/2023/formulas/mathematics/college/djbtddhaaylhmasygclp8yhc51s3uy8s26.png)
We take the arc tan of both sides.
![\tan( - 1) ( \tan( (\pi)/(3)x - 3 ) = \tan {}^( - 1) ( {}^( )0 )](https://img.qammunity.org/2023/formulas/mathematics/college/yehycc7jvwfosztbcxo63hh45n658f6i2k.png)
![(\pi)/(3) x - 3 = 0](https://img.qammunity.org/2023/formulas/mathematics/college/46xp8xzpjklzxg0vp4pzjijeqmtjoehz5b.png)
![(\pi)/(3) x = 3](https://img.qammunity.org/2023/formulas/mathematics/college/9ipet7j14da2zrj2yuttdtc8a6537vp536.png)
![= (9)/(\pi)](https://img.qammunity.org/2023/formulas/mathematics/college/xcnjw1daux47ql7bwpcbifxzgfus0jo909.png)
Since the period of a tan function, is pi, we divide pi by pi/3 since pi/3is the coeffeicent of the x variable
![(\pi)/( (\pi)/(3) ) = 3](https://img.qammunity.org/2023/formulas/mathematics/college/3xsakj6i104n9ubvkalsbg9w5tgx4iyl8h.png)
So the answer is
![(9)/(\pi) + 3n](https://img.qammunity.org/2023/formulas/mathematics/college/4f3n8r3jpfo3o8h3nh3ohijhbww2e2g9mv.png)
If this the question,
![\tan( (\pi)/(3) x) = 3](https://img.qammunity.org/2023/formulas/mathematics/college/bvjbhzvhptll2yinzjpwiqzfeyz9n8rux6.png)
![(\pi)/(3) x = 1.249](https://img.qammunity.org/2023/formulas/mathematics/college/79e38q77ozg7s61j19d11vio2z1esgn4sb.png)
![x = (3.747)/(\pi)](https://img.qammunity.org/2023/formulas/mathematics/college/rnmekckggzuq8xy0b74tquy56qpld3etk5.png)
The period is once again 3 so we have
![(3.747)/(\pi) + 3n](https://img.qammunity.org/2023/formulas/mathematics/college/72yux52huzrnmdeja9098k34ay65lni0ch.png)
where n is a interger.