Chapter 7.3, Problem 16AYU

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

$\sqrt{3}\cot \theta +1=0$

To determine

To find: A linear trigonometric equation $\surd 3\text{cot}\theta +1=0$ for $\theta$.

Answer to Problem 16AYU

The solution set $=\left\{2\pi /3,5\pi /3\right\}$.

Explanation of Solution

Given:

A linear trigonometric equation $\surd 3\text{cot}\theta +1=0$.

Calculation:

To find $\theta$, we need to solve for $\text{cot}\theta$.

$\surd 3\text{cot}\theta +1=0$

Subtract 1 on both sides.

$\surd 3\text{cot}\theta =-1$

Dividing $\sqrt{3}$ on both sides,

$\text{cot}\theta =-1/\surd 3$

Using the formula,

$\text{cot}\theta =1/\text{tan}\theta$

$1/\text{tan}\theta =-1/\surd 3$

$\text{tan}\theta =-\surd 3$

In the interval $0\le \theta <2\pi$, or in the interval notation, $\left[0,2\pi \right)$, we need to find angles for which $\text{tan}\theta =-\surd 3$.

$\text{θ}=2\pi /3,5\pi /3$

Therefore, the solution set $=\left\{2\pi /3,5\pi /3\right\}$.