Solving Trig Equations: Understanding Degrees and Terminal Arm Usage

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Homework Help Overview

The discussion revolves around solving trigonometric equations, specifically focusing on the equations involving sine and cosine functions. Participants express confusion about determining the angles in degrees and the use of terminal arms in the context of these equations.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants attempt to solve equations by isolating sine and cosine terms and using inverse functions to find angles. Questions arise about how to determine the correct angles and the relevance of the terminal arm concept in predicting these angles.

Discussion Status

Some participants provide partial guidance on the approach to take, including checking values on calculators and considering the quadrants where sine and cosine are positive. There is an ongoing exploration of different interpretations of the equations and the methods to solve them.

Contextual Notes

There is mention of assumptions regarding the range of angles (0 to 2π) and a suggestion to clarify notation for square roots. Some participants express uncertainty about the foundational geometric concepts that may be necessary for understanding the problem.

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Homework Statement



I don't understand how to solve for these, I will post 3.

d) [sqrt]2sin x + 1 = 0 (sqrt only over the 2)
e) 2cos x - [sqrt]3 = 0
f) 2sinx +[sqr]3 = 0

Esentially, I don't understand how to get the degrees. I don't get how you use the terminal arm to predict the degrees, etc.

Can anyone help?



Homework Equations





The Attempt at a Solution



1 / [sqrt]2

sin inverse of that is 30 degrees.

which is pi over 6.

How do i get the second one? how do i know? etc
 
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Nelo said:
d) [sqrt]2sin x + 1 = 0 (sqrt only over the 2)
e) 2cos x - [sqrt]3 = 0
f) 2sinx +[sqr]3 = 0

I am assuming you are finding from 0 to 2pi.

Nelo said:

The Attempt at a Solution



1 / [sqrt]2

sin inverse of that is 30 degrees.

which is pi over 6.

How do i get the second one? how do i know? etc

For the three you are doing the same thing, making sinx or cosx the subject of the formula and then taking the inverse function of it.

The first one you did partially right

sinx = 1/√2 and x=sin-1(1/√2)

Check back on your calculator for what value this is.

When you get that what quadrant is sine positive in?
 
Nelo said:

Homework Statement



I don't understand how to solve for these, I will post 3.

d) [sqrt]2sin x + 1 = 0 (sqrt only over the 2)
e) 2cos x - [sqrt]3 = 0
f) 2sinx +[sqr]3 = 0

Esentially, I don't understand how to get the degrees. I don't get how you use the terminal arm to predict the degrees, etc.

Can anyone help?



Homework Equations





The Attempt at a Solution



1 / [sqrt]2

sin inverse of that is 30 degrees.

which is pi over 6.

How do i get the second one? how do i know? etc

Write sqrt(2) instead of [sqrt]2, etc; that way, you don't have to make statements like "sqrt only over the 2". Anyway, if you know elementary Geometry, you should be able to see what angles give you sin(x) = -1/sqrt(2), cos(x) = 3/sqrt(2) and sin(x) = -sqrt(3)/2. I realize that Geometry may not be taught anymore in schools, in which case you can do a Google search on something like "sines of special angles" to find some material.

RGV
 
4f) cosx =2sincosx

How do you solve this?

The only thing i see is

..cosx - cosx- = 2sin

is that wrong?

does that mean cos is 0 and sin is 90+90 = 180 degrees?
 
Nelo said:
4f) cosx =2sincosx

How do you solve this?

The only thing i see is

..cosx - cosx- = 2sin

is that wrong?

does that mean cos is 0 and sin is 90+90 = 180 degrees?

bring the cosx on the left side to the right side and then factor out the cosx.
 
Does that not equal the exact same thing? cos is 0 and sin is 180?
 
Nelo said:
Does that not equal the exact same thing? cos is 0 and sin is 180?

You mean x=0, x=180? if so, by putting x=0 into the equation you will see that it is not a solution.
 

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