Chapter 7.3, Problem 28AYU

In Problems 13-36, solve each equation on the interval $0\le \theta <2\pi$.

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

To determine

To find: The trigonometric equation $\text{tan}\left(2\theta \right)=-1$, between $0\le \theta <2\pi$.

Answer to Problem 28AYU

The solution set for the trigonometric equation $\text{tan}\left(2\theta \right)=-1=\{-\pi /8,3\pi /8,7\pi /8,11\pi /8\}$ for $0\le \theta <2\pi$.

Explanation of Solution

Given:

Trigonometric equation $\text{tan}\left(2\theta \right)=-1$, between $0\le \theta <2\pi$.

Calculation:

The angle $\theta$ for which $\text{tan}$ function has a value of $-1=3\pi /4,4\pi /3$.

Therefore, the graph $y=\text{ tan}\left(2\theta \right)$ intersects the graph of $y=-1$ at $3\pi /4,7\pi /4$.

For the interval $\left[0,2\pi \right)$.

The period of $y=\tan \left(2\theta \right)$.

$2\theta =\pi$

The general formula of $\text{tan}\left(2\theta \right)=-1$.

$2\theta =3\pi /4+\pi k$ or $2\theta =7\pi /4+\pi k$, where $k$ is any integer.

Dividing by 2 on each side,

$\theta =3\pi /8+\pi k/2$ or $\theta =7\pi /8+\pi k/2$

For different values of $k$:

1. When $k=-1$.

$\theta =3\pi /8+\pi \left(-1\right)/2$ or $\theta =7\pi /8+\pi \left(-1\right)/2$

$\theta =3\pi /8-\pi /2=-\pi /8$ or $\theta =7\pi /8-\pi /2=3\pi /8$

2. When $k=0$.

$\theta =3\pi /8+\pi \left(0\right)/2$ or $\theta =7\pi /8+\pi \left(0\right)/2$

$\theta =3\pi /8$ or $\theta =7\pi /8$

3. When $k=\text{ 1}$.

$\theta =3\pi /8+\pi \left(1\right)/2$ or $\theta =7\pi /8+\pi \left(1\right)/2$

$\theta =3\pi /8+\pi /2=7\pi /8$ or $\theta =7\pi /8+\pi /2=11\pi /8$

The solution set for the trigonometric equation $\text{tan}\left(2\theta \right)=-1$ for the interval $\left[0,2\pi \right)=\{-\pi /8,3\pi /8,7\pi /8,11\pi /8\}$ for $0\le \theta <2\pi$.