Chapter 7.3, Problem 30AYU

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

$\cot \frac{2\theta }{3}=-\sqrt{3}$

To determine

To find: The trigonometric equation $\text{cot}\left(2\text{θ}/3\right)=-\sqrt{3}$, between $0\le \theta <2\pi$.

Answer to Problem 30AYU

The solution set for the trigonometric equation $\text{cot}\left(2\text{θ}/3\right)=-\sqrt{3}=\left\{-\pi /4,5\pi /4,11\pi /4,7\pi /4\right\}$. $0\le \theta <2\pi$.

Explanation of Solution

Given:

Trigonometric equation $\text{cot}\left(2\text{θ}/3\right)=-\sqrt{3}$, $0\le \theta <2\pi$.

Calculation:

Using the formula,

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

$\text{cot}\left(2\text{θ}/3\right)=-\sqrt{3}=1/\text{tan}\left(2\theta /3\right)$

$\text{tan}\left(2\theta /3\right)=-1/\sqrt{3}$

The angle $\theta$ for which $\text{tan}$ function has a value of $-1/\sqrt{3}=5\pi /6,11\pi /6$.

Therefore, the graph $y=\text{tan}\left(2\theta /3\right)$ intersects the graph of $y=-1/\sqrt{3}$ at $5\pi /6$, $11\pi /6$.

$\Theta$ will be in II and IV quadrants.

But there will be other angles for which $\text{tan}\left(2\theta /3\right)=-1/\sqrt{3}$.

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

The period of $y=\text{tan}\left(2\theta /3\right)$.

$2\theta /3=\pi$

The general formula of $\text{tan}\left(2\theta /3\right)=-1/\sqrt{3}$.

$2\theta /3=5\pi /6+\pi k$ or $2\theta /3=11\pi /6+\pi k$, where $k$ is any integer.

Multiplying each side by 3 and dividing each side by 2,

$\Theta =15\pi /12+3\pi k/2\text{ or }\theta =33\pi /12+3\pi k/2$

For different values of $k$:

1) When $k=-1$

$\Theta =15\pi /12+3\pi \left(-1\right)/2\text{ or }\theta =33\pi /12+3\pi \left(-1\right)/2$

$\Theta =15\pi /12-3\pi /2\text{ or }\theta =33\pi /12-3\pi /2$

$\Theta =-\pi /4\text{ or }\theta =5\pi /4$

2) When $k=0$

$\Theta =15\pi /12+3\pi \left(0\right)/2\text{ or }\theta =33\pi /12+3\pi \left(0\right)/2$

$\Theta =5\pi /4\text{ or }\theta =11\pi /4$

3) When $k=-2$

$\Theta =15\pi /12+3\pi \left(-2\right)/2\text{ or }\theta =33\pi /12+3\pi \left(-2\right)/2$

$\Theta =15\pi /12-6\pi /2\text{ or }\theta =33\pi /12-6\pi /2$

$\Theta =-7\pi /4\text{ or }\theta =-\pi /4$

The solution set for the trigonometric equation $\text{sec}\left(3\theta /2\right)=-2$ for the interval $\left[0,2\pi \right)=\left\{-\pi /4,5\pi /4,11\pi /4,7\pi /4\right\}$. $0\le \theta <2\pi$.