Solving Equations Using the Unit Circle

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Homework Help Overview

The discussion revolves around finding all solutions to the equation 2sin2x + sinx = 0 within the interval -180° ≤ x ≤ 90°. Participants are exploring the use of the unit circle to derive exact values for the sine function.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are attempting to solve the equation by factoring and considering the unit circle for exact values. Questions arise regarding the solutions for sin(x) = -1/2 and the implications of the specified interval.

Discussion Status

There is an ongoing exploration of the solutions to the equation, with some participants noting the need to adhere to the interval constraints. Guidance is offered regarding the interpretation of sine values and the relevance of the unit circle.

Contextual Notes

Participants are reminded of the interval -180° ≤ x ≤ 90°, which limits the acceptable solutions. There is also mention of potential typos in the equations presented.

kylepetten
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Homework Statement



Find all solutions to the equation below such that -180° [tex]\leq[/tex] x [tex]\leq[/tex] 90°

2sin2x + sinx = 0

Homework Equations


The Attempt at a Solution



2sin2x + sinx = 0

sinx[2sinx + 1] = 0

sin x = 0 sinx= -1/2

x = {0° + 360°n
{-180° + 360°n
n[tex]\epsilon[/tex]I
 
Last edited:
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kylepetten said:

Homework Statement



Find all solutions to the equation below such that -180° [tex]\leq[/tex] x [tex]\leq[/tex] 90°

2sin2x + sinx = 0


Homework Equations





The Attempt at a Solution



2sin2x + sinx = 0

sinx[2sinx + 1] = 0

sin x = 0 sinx= -1/2

x = {0° + 360°n
{-180° + 360°n
n[tex]\epsilon[/tex]I

x = n*pi are the solutions to sin(x) = 0. What about the solutions to sin(x) = -1/2?
 
Mark44 said:
x = n*pi are the solutions to sin(x) = 0. What about the solutions to sin(x) = -1/2?

im not using equations, i am getting my exact values from the unit circle
 
kylepetten said:
im not using equations, i am getting my exact values from the unit circle
Yeah, Mark44! :smile:
 
kylepetten said:
im not using equations, i am getting my exact values from the unit circle
You are using equations, namely 2sin2x + sinx = 0, which you rewrote as sinx(2sinx + 1) = 0.

You have found the solutions to the equation sinx = 0, but you haven't found any for the equation 2sinx + 1 = 0.

BTW, according to your problem description, you need be concerned only with values for which -180° <= x <= 90°. On this interval there are only two solutions to sinx = 0.
 
kylepetten said:
im not using equations, i am getting my exact values from the unit circle

If you post an equation here, it looks like you're using it.
sin x = 1/2 is another possibility. I'm certain that value of x is on the unit circle as well.
you didn't use the condition that -180<=x<=90. An answer such as 360n isn't acceptable here.
 
willem2 said:
If you post an equation here, it looks like you're using it.
sin x = 1/2 is another possibility. I'm certain that value of x is on the unit circle as well.
you didn't use the condition that -180<=x<=90. An answer such as 360n isn't acceptable here.
Probably a typo, but sin x = 1/2 is not a solution. sin x = -1/2 is a solution, though.
 

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