Question

In Problems 11–34, solve each equation on the interval $0\le \theta \le 2\pi$

$\tan \left(2\theta \right)=-1$

Short answer

The solution set is$\left\{\frac{3\mathrm{\pi }}{8},\frac{7\mathrm{\pi }}{8},\frac{11\mathrm{\pi }}{8},\frac{15\mathrm{\pi }}{8}\right\}$.

Step-by-step solution

Step 1. Given Information 

In the given problem we have to solve each equation on the interval$0\le \theta \le 2\pi$

$\tan \left(2\theta \right)=-1$

Step 2. In the interval [0,2π), the tangent function equals -1 at 3π4

$2\theta$must equal to $\frac{3\mathrm{\pi }}{4}$.

$2\theta =\frac{3\mathrm{\pi }}{4}+\pi n$

Divide by 2 on both side

$$\begin{gathered} \frac{2}{2}\theta =\frac{1}{2}\left(\frac{3\mathrm{\pi }}{4}+\pi n\right) \\ \theta =\frac{1}{2}·\frac{3\mathrm{\pi }}{4}+\frac{1}{2}·\pi n \\ \theta =\frac{3\mathrm{\pi }}{8}+\frac{\pi n}{2} \end{gathered}$$

Step 3. The general formula is θ=3π8+πn2

So the value of given function in interval $[0,2\mathrm{\pi })$is

$$\begin{gathered} \theta =\frac{3\mathrm{\pi }}{8}+\frac{\pi \times 0}{2}\theta =\frac{3\mathrm{\pi }}{8}+\frac{\pi \times 1}{2}\theta =\frac{3\mathrm{\pi }}{8}+\frac{\pi \times 2}{2}\theta =\frac{3\mathrm{\pi }}{8}+\frac{\pi \times 3}{2} \\ \theta =\frac{3\mathrm{\pi }}{8}\theta =\frac{3\mathrm{\pi }}{8}+\frac{\pi }{2}\theta =\frac{3\mathrm{\pi }}{8}+\pi \theta =\frac{3\mathrm{\pi }}{8}+\frac{3\pi }{2} \\ \theta =\frac{3\mathrm{\pi }}{8}\theta =\frac{3\mathrm{\pi }+4\mathrm{\pi }}{8}\theta =\frac{3\mathrm{\pi }+8\mathrm{\pi }}{8}\theta =\frac{3\mathrm{\pi }+12\mathrm{\pi }}{8} \\ \theta =\frac{3\mathrm{\pi }}{8}\theta =\frac{7\mathrm{\pi }}{8}\theta =\frac{11\mathrm{\pi }}{8}\theta =\frac{15\mathrm{\pi }}{8} \end{gathered}$$

So the solution set is$\left\{\frac{3\mathrm{\pi }}{8},\frac{7\mathrm{\pi }}{8},\frac{11\mathrm{\pi }}{8},\frac{15\mathrm{\pi }}{8}\right\}$.