Chapter 4.4, Problem 96E

(a)

To determine

Two solutions of $\cot \theta =0$ in radians $0\le \theta <2\pi$ and in degrees $0^0\le \theta <{360}^0$ .

Answer to Problem 96E

  $\theta =\frac{\pi }{2}={90}^0\text{ and }\theta =\frac{3\pi }{2}={270}^0$

Explanation of Solution

Given info:

  $\cot \theta =0$

Formula used:

In quadrant I all trigonometric functions are positive.

In quadrant II sine functions are positive.

In quadrant III tangent functions are positive.

In quadrant IV cosine functions are positive.

Calculation:

Given that $\cot \theta =0$ ,

We have

  $\begin{array}{l}\cot \theta =0\\ \theta =\frac{\pi }{2}={90}^0\text{ and }\theta =\frac{3\pi }{2}={270}^0\end{array}$

Conclusion:

Thus,the solutions are $\theta =\frac{\pi }{2}={90}^0\text{ and }\theta =\frac{3\pi }{2}={270}^0$

(b)

To determine

Two solutions of $\sec \theta =-\sqrt{2}$ in radians $0\le \theta <2\pi$ and in degrees $0^0\le \theta <{360}^0$ .

Answer to Problem 96E

  $\theta =\frac{3\pi }{4}={135}^0\text{ and }\theta =\frac{5\pi }{4}={225}^0$

Explanation of Solution

Given info:

  $\sec \theta =-\sqrt{2}$

Formula used:

In quadrant I all trigonometric functions are positive.

In quadrant II sine functions are positive.

In quadrant III tangent functions are positive.

In quadrant IV cosine functions are positive.

Calculation:

Given that $\sec \theta =-\sqrt{2}$ ,

We have

  $\begin{array}{l}\sec \theta =-\sqrt{2}\\ \theta =\frac{3\pi }{4}={135}^0\text{ and }\theta =\frac{5\pi }{4}={225}^0\end{array}$

Conclusion:

Thus,the solutions are $\theta =\frac{3\pi }{4}={135}^0\text{ and }\theta =\frac{5\pi }{4}={225}^0$