Chapter 7.3, Problem 51AYU

In Problems 47-58, use a calculator to solve each equation on the interval $0\le \theta <2\pi$. Round answers to two decimal places.

$\sec \theta =-4$

To determine

To find: The trigonometric equation $\text{sec}\theta =-4$ using the calculator, and rounding off the answer to two decimal places, $0\le \theta <2\pi$.

Answer to Problem 51AYU

The solution set for the trigonometric equation $\text{sec}\theta =-4=\left\{{1.82}^c,{4.46}^c\right\},0\le \theta <2\pi$.

Explanation of Solution

Given:

Trigonometric equation $\text{sec}\theta =-4$, $0\le \theta <2\pi$.

Calculation:

To solve $\text{sec}\theta =-4$ on a calculator, the mode has to be kept in radians mode. Here the function is $\text{sec}$.

Therefore, to calculate for $\theta$ we cannot directly use the calculator.

Using the formula

$\text{Sec}\theta =1/\text{cos}\theta$

$1/\text{cos}\theta =-4$

$\text{cos}\theta =-0.25$

$\Theta ={\text{ cos}}^{-1}\left(-0.25\right)\approx 1.823$

Rounding off to two decimal places,

$\Theta ={\text{ cos}}^{-1}\left(-0.25\right)=1.82$ radians.

By definition,

$\theta ={\text{ cos}}^{-1}x$, the angle $\theta$ is defined within the range as $0\le \theta \le \pi$.

For $\text{cos}\theta =-0.25$, we need to find solutions for $\theta$ for $0\le \theta <2\pi$ interval.

$\text{cos}\theta =-0.25<0$

Therefore,

$\Theta$ lies in II and III quadrants.

II quadrant angle $\theta ={1.82}^c$.

III quadrant angle $\theta =2\pi -1.82={4.46}^c$.

The solution set for Trigonometric equation $\text{sec}\theta =-4={1.82}^c,{4.46}^c$.