Question

1. find all radian solutions to the equation (tan(3x))² = 1.

Explanation:

Step1: Take square roots of both sides

Given \((\tan(3x))^2 = 1\), taking square roots on both sides, we get \(\tan(3x)=\pm1\).

Step2: Solve \(\tan(3x) = 1\)

We know that \(\tan\theta = 1\) when \(\theta=\frac{\pi}{4}+n\pi\), where \(n\in\mathbb{Z}\). So for \(\tan(3x) = 1\), we have \(3x=\frac{\pi}{4}+n\pi\). Dividing both sides by 3, we get \(x = \frac{\pi}{12}+\frac{n\pi}{3}\), where \(n\in\mathbb{Z}\).

Step3: Solve \(\tan(3x)=- 1\)

We know that \(\tan\theta=-1\) when \(\theta = \frac{3\pi}{4}+n\pi\), where \(n\in\mathbb{Z}\). So for \(\tan(3x)=-1\), we have \(3x=\frac{3\pi}{4}+n\pi\). Dividing both sides by 3, we get \(x=\frac{\pi}{4}+\frac{n\pi}{3}\), where \(n\in\mathbb{Z}\).

Step4: Combine the solutions

The solutions from \(\tan(3x) = 1\) and \(\tan(3x)=-1\) can be combined. Notice that \(\frac{\pi}{12}+\frac{n\pi}{3}=\frac{\pi + 4n\pi}{12}=\frac{(4n + 1)\pi}{12}\) and \(\frac{\pi}{4}+\frac{n\pi}{3}=\frac{3\pi+4n\pi}{12}=\frac{(4n + 3)\pi}{12}\). We can also write the general solution as \(x=\frac{\pi}{12}+\frac{n\pi}{6}\), where \(n\in\mathbb{Z}\). Let's check:
If \(n = 2k\) (even), \(x=\frac{\pi}{12}+\frac{2k\pi}{6}=\frac{\pi}{12}+\frac{k\pi}{3}=\frac{(4k + 1)\pi}{12}\) (same as from \(\tan(3x)=1\)).
If \(n=2k + 1\) (odd), \(x=\frac{\pi}{12}+\frac{(2k + 1)\pi}{6}=\frac{\pi+2\pi + 2k\pi}{12}=\frac{(4k + 3)\pi}{12}\) (same as from \(\tan(3x)=-1\)).

Another way: From \(3x=\frac{\pi}{4}+n\pi\) or \(3x=\frac{3\pi}{4}+n\pi\), dividing by 3:
For \(3x=\frac{\pi}{4}+n\pi\), \(x=\frac{\pi}{12}+\frac{n\pi}{3}\); for \(3x=\frac{3\pi}{4}+n\pi\), \(x=\frac{\pi}{4}+\frac{n\pi}{3}=\frac{3\pi}{12}+\frac{4n\pi}{12}=\frac{(4n + 3)\pi}{12}\) and \(x=\frac{\pi}{12}+\frac{4n\pi}{12}=\frac{(4n+1)\pi}{12}\). We can rewrite the general solution as \(x=\frac{\pi}{12}+\frac{n\pi}{6}\), \(n\in\mathbb{Z}\) (since \(\frac{n\pi}{3}=\frac{2n\pi}{6}\), and combining \(\frac{\pi}{12}+\frac{2n\pi}{6}\) and \(\frac{\pi}{12}+\frac{(2n + 1)\pi}{6}\) gives \(\frac{\pi}{12}+\frac{n\pi}{6}\) for all integers \(n\)).

Answer:

The solutions are \(x=\frac{\pi}{12}+\frac{n\pi}{6}\), where \(n\) is any integer (\(n\in\mathbb{Z}\)).