Trigonometric Equation: Solving for Sec x=2 and Reviewing for a Test

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SUMMARY

The discussion focuses on solving the trigonometric equation sec(4x) = 2, leading to the solutions for sec(x) = 2, which are x = π/3 and x = 5π/3. The user seeks clarification on how to derive these angles and their corresponding values when divided by 4, resulting in x = π/12 and x = 5π/12. The explanation emphasizes the use of the unit circle and the properties of cosine to find all possible solutions, including the periodic nature of trigonometric functions.

PREREQUISITES
  • Understanding of trigonometric functions, specifically secant and cosine.
  • Familiarity with the unit circle and angle measurement in radians.
  • Knowledge of inverse trigonometric functions, particularly arccosine.
  • Basic concepts of periodicity in trigonometric equations.
NEXT STEPS
  • Study the properties of the secant function and its relationship with cosine.
  • Learn how to solve trigonometric equations involving multiple angles, such as sec(4x).
  • Explore the unit circle in detail, focusing on how to derive angles from cosine values.
  • Investigate the periodicity of trigonometric functions and how to express general solutions.
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Students preparing for trigonometry tests, educators teaching trigonometric equations, and anyone looking to deepen their understanding of secant and cosine functions.

Phyzwizz
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This is a problem from a quiz I had awhile ago, I got the answer correct, but I'm reviewing for a test and I completely forget how I did it. Can someone please show me what I did?

sec4x=2
secx=2
x=π/3 + 2nπ ---> x=π/12+2nπ/4
x=5π/3 + 2nπ ---> x=5π/12+2nπ/4

π=pi

p.s. I mostly just forget how to get the value for sec x= 2 how did I get π/3 and 5π/3

Thanks!
 
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Phyzwizz said:
This is a problem from a quiz I had awhile ago, I got the answer correct, but I'm reviewing for a test and I completely forget how I did it. Can someone please show me what I did?

sec4x=2
secx=2
x=π/3 + 2nπ ---> x=π/12+2nπ/4
x=5π/3 + 2nπ ---> x=5π/12+2nπ/4

π=pi

p.s. I mostly just forget how to get the value for sec x= 2 how did I get π/3 and 5π/3

Thanks!

First off, I have no idea why you have two problems written right after each other. I will assume you sought to solve the first.
[tex]sec(4x) = 2 \to \frac{1}{2} = cosine(4x) \to x = \frac{arccos(\frac{1}{2})}{4}[/tex]

So you are taking an inverse cosine of one half. So the question is, "The cosine of what angle gives you 1/2?"

To achieve this, draw the unit circle. Remember that cosinusoids correspond to the x-axis, and sinusoids correspond to the y-axis. Thus, this positive answer of 1/2 must correspond to regions in the graph where x are positive. These regions are I and IV, using the common numbering system. You then draw the two 30-60-90 degree triangles, label the side of both triangles that correspond to the x-axis "1/2," label the hypotenuse as 1, and label the remaining side sqrt(3)/2. Then, use your triangular knowledge to figure out the correct angle.
 
You solved sec(x)=2 ...which gives you Pi/3 and 5*Pi/3, but you added 2*Pi*n because you can continually go around the unit circle and find answers. After your ----> thing, you divided both by 4 in order to come up with x=arcsec(2)/4, which is what tedbradly has shown you.

The matter is choosing all angles in one time around the unit circle that fits with this. So far, you have answers (just your answers for sec(x)=2, each divided by 4):

Pi/12
5*Pi/12

But you can also move around the unit circle by n*Pi/2 to find other answers. n is always an integer, since it has not changed from the 2*Pi*n case .

So you have Pi/12, Pi/12+Pi/2, Pi/12+Pi, Pi/12+3Pi/2...5*Pi/12, 5*Pi/12+Pi/2, 5*Pi/12+Pi, 5*Pi/12+3Pi/2.

This makes sense because you have 2 answers initially and you were solving for 4x...4*2=8 solns.
 

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