Chapter 4.4, Problem 106E

(a)

To determine

To calculate:The two solutions of the equation $\cot \theta =-\sqrt{3}$ in both degrees and radians.

Answer to Problem 106E

The two solutions of the equation $\cot \theta =-\sqrt{3}$ are $\theta =150°,330°$ , and in radians $\theta =\frac{5\pi }{6},\frac{11\pi }{6}$ .

Explanation of Solution

Given:

The trigonometric equation is $\cot \theta =-\sqrt{3}$ and $0°\le \theta <360°$ or $0\le \theta <2\pi$ in radians.

Calculation:

Since $\cot 30°=\sqrt{3}$ and $\cot \left(-\theta \right)=-\cot \theta$ , therefore $\cot \left(-30°\right)=-\sqrt{3}$ , therefore $\theta =-30°$ .

As per the trigonometric identity $\cot \left(180°-\theta \right)=-\cot \theta$ , as $\cot \left(180°-\theta \right)=-\cot 30°$ , this gives

  $\begin{array}{c}\theta =180°-30°\\ =150°\end{array}$

Also, $\cot \left(360°-\theta \right)=-\cot \theta$ , as $\cot \left(360°-\theta \right)=-\cot 30°$ , this gives

  $\begin{array}{c}\theta =360°-30°\\ =330°\end{array}$ .

Convert $\theta =150°$ in radians.

  $\begin{array}{c}\theta =150°\times \frac{\pi }{180°}\\ =\frac{5\pi }{6}\end{array}$

For $\theta =330°$ in radians.

  $\begin{array}{c}\theta =330°\times \frac{\pi }{180°}\\ =\frac{11\pi }{6}\end{array}$ .

Therefore the values of $\theta$ for $\cot \theta =-\sqrt{3}$ is $\theta =150°,330°$ in degrees and $\theta =\frac{5\pi }{6},\frac{11\pi }{6}$ in radians.

(b)

To determine

To calculate:The two solutions of the equation $\csc \theta =2$ in both degrees and radians.

Answer to Problem 106E

The two solutions of the equation $\csc \theta =2$ are $\theta =30°,150°$ , and in radians $\theta =\frac{\pi }{6},\frac{5\pi }{6}$ .

Explanation of Solution

Given:

The trigonometric equation is $\csc \theta =2$ and $0°\le \theta <360°$ or $0\le \theta <2\pi$ in radians.

Calculation:

Since, the value of cosecant is $2$ and it gives the internal angles are $30°,90°$ and $60°$ . The opposite side is $1$ and the hypotenuse is $2$ . The value of base is $\sqrt{3}$ . Draw a triangle

  

So, the angle $\theta =30°$ and as per the trigonometric identity $\csc \left(180°-\theta \right)=\csc \theta$ , therefore another value of

  $\begin{array}{c}\theta =180°-30°\\ =150°\end{array}$ .

Convert $\theta =30°$ in radians.

  $\begin{array}{c}\theta =30°\times \frac{\pi }{180°}\\ =\frac{\pi }{6}\end{array}$

For $\theta =150°$ in radians.

  $\begin{array}{c}\theta =150°\times \frac{\pi }{180°}\\ =\frac{5\pi }{6}\end{array}$

Therefore the values of $\theta$ for $\csc \theta =2$ is $\theta =30°,150°$ in degrees and $\theta =\frac{\pi }{6},\frac{5\pi }{6}$ in radians.