Question

solve the equation for exact solutions over the interval (0, 2π).
6 tan 3x = 6
select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the solution set is
(type an exact answer, using π as needed. type your answer in radians. use integers or fractions for any numbers in the expression. use a comma to separate answers as needed)
b. the solution is the empty set

Explanation:

Step1: Simplify the equation

Divide both sides of the equation \(6\tan 3x = 6\) by 6.
\(\tan 3x=\frac{6}{6} = 1\)

Step2: Find the general solution for \(3x\)

We know that \(\tan\theta = 1\) when \(\theta=\frac{\pi}{4}+n\pi\), where \(n\) is an integer. So for \(\tan 3x = 1\), we have \(3x=\frac{\pi}{4}+n\pi\)

Step3: Solve for \(x\)

Divide both sides of the equation \(3x=\frac{\pi}{4}+n\pi\) by 3.
\(x = \frac{\pi}{12}+\frac{n\pi}{3}\)

Step4: Find the solutions in the interval \((0, 2\pi)\)

We need to find all integer values of \(n\) such that \(0<\frac{\pi}{12}+\frac{n\pi}{3}<2\pi\)

• For \(n = 0\): \(x=\frac{\pi}{12}\), and \(0<\frac{\pi}{12}<2\pi\) (since \(\frac{\pi}{12}\approx0.26\) and \(2\pi\approx6.28\))
• For \(n = 1\): \(x=\frac{\pi}{12}+\frac{\pi}{3}=\frac{\pi + 4\pi}{12}=\frac{5\pi}{12}\), and \(0<\frac{5\pi}{12}<2\pi\) ( \(\frac{5\pi}{12}\approx1.31\))
• For \(n = 2\): \(x=\frac{\pi}{12}+\frac{2\pi}{3}=\frac{\pi+ 8\pi}{12}=\frac{9\pi}{12}=\frac{3\pi}{4}\), and \(0<\frac{3\pi}{4}<2\pi\) ( \(\frac{3\pi}{4}\approx2.36\))
• For \(n = 3\): \(x=\frac{\pi}{12}+\pi=\frac{\pi + 12\pi}{12}=\frac{13\pi}{12}\), and \(0<\frac{13\pi}{12}<2\pi\) ( \(\frac{13\pi}{12}\approx3.40\))
• For \(n = 4\): \(x=\frac{\pi}{12}+\frac{4\pi}{3}=\frac{\pi+16\pi}{12}=\frac{17\pi}{12}\), and \(0<\frac{17\pi}{12}<2\pi\) ( \(\frac{17\pi}{12}\approx4.45\))
• For \(n = 5\): \(x=\frac{\pi}{12}+\frac{5\pi}{3}=\frac{\pi + 20\pi}{12}=\frac{21\pi}{12}=\frac{7\pi}{4}\), and \(0<\frac{7\pi}{4}<2\pi\) ( \(\frac{7\pi}{4}\approx5.49\))
• For \(n = 6\): \(x=\frac{\pi}{12}+2\pi=\frac{\pi + 24\pi}{12}=\frac{25\pi}{12}\), and \(0<\frac{25\pi}{12}<2\pi\) ( \(\frac{25\pi}{12}\approx6.54\))
• For \(n = 7\): \(x=\frac{\pi}{12}+\frac{7\pi}{3}=\frac{\pi+28\pi}{12}=\frac{29\pi}{12}\), and \(\frac{29\pi}{12}\approx7.59>2\pi\) (since \(2\pi=\frac{24\pi}{12}\)), so \(n = 7\) is too big.
• For \(n=- 1\): \(x=\frac{\pi}{12}-\frac{\pi}{3}=\frac{\pi - 4\pi}{12}=-\frac{3\pi}{12}=-\frac{\pi}{4}\), and \(-\frac{\pi}{4}<0\), so it is not in the interval \((0, 2\pi)\)

So the solutions are \(\frac{\pi}{12},\frac{5\pi}{12},\frac{3\pi}{4},\frac{13\pi}{12},\frac{17\pi}{12},\frac{7\pi}{4},\frac{25\pi}{12}\)

Answer:

A. The solution set is \(\frac{\pi}{12},\frac{5\pi}{12},\frac{3\pi}{4},\frac{13\pi}{12},\frac{17\pi}{12},\frac{7\pi}{4},\frac{25\pi}{12}\)