Chapter 14.2, Problem 38PPSE

Solution

To Solve: $\sec \theta =2$for $0\le \theta <2\pi$ .

  $\theta =\frac{\pi }{3}\text{ or }\theta =\frac{5\pi }{3}$for $0\le \theta <2\pi$ .

Given information:

The equation $\sec \theta =2$

Concept and Formula Used:

To solve the above expression, use the values of inverse trigonometric ratios on the unit circle. Also, secant function is positive in the first and fourth quadrant. So, the solution will be $x,2\pi -x$ .

  $\begin{array}{l}\left(i\right){\sec }^{-1}2=\frac{\pi }{3},\frac{5\pi }{3}\\ \left(ii\right){\sec }^{-1}\left(\sec x\right)=x\end{array}$

Calculation:

The given equation is $\sec \theta =2$.

Taking inverse of secant function both the sides:

  $\begin{array}{c}{\sec }^{-1}\left(\sec \theta \right)={\sec }^{-1}\left(2\right)\\ \theta ={\sec }^{-1}\left(2\right)\\ \theta =\frac{\pi }{3},2\pi -\frac{\pi }{3}=\frac{5\pi }{3}\end{array}$

Conclusion:

  $\theta =\frac{\pi }{3}\text{ or }\theta =\frac{5\pi }{3}$for $0\le \theta <2\pi$ .