Solving Equations Using the Unit Circle

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SUMMARY

The discussion focuses on solving the equation 2sin2x + sinx = 0 within the interval -180° ≤ x ≤ 90°. The solutions identified include x = 0° and x = -180°, derived from sinx = 0, and x = -30° from sinx = -1/2. Participants emphasize using the unit circle for exact values, clarifying that solutions must adhere to the specified interval. The conversation highlights the importance of correctly interpreting the conditions of the problem.

PREREQUISITES
  • Understanding of trigonometric functions, specifically sine.
  • Familiarity with the unit circle and its values.
  • Knowledge of solving trigonometric equations.
  • Ability to interpret solution intervals in trigonometric contexts.
NEXT STEPS
  • Study the unit circle for sine values and their corresponding angles.
  • Learn how to solve trigonometric equations using identities.
  • Explore the implications of solution intervals in trigonometric problems.
  • Practice solving similar equations, such as 2cos2x + cosx = 0.
USEFUL FOR

Students studying trigonometry, educators teaching trigonometric equations, and anyone seeking to enhance their understanding of the unit circle and its applications in solving equations.

kylepetten
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Homework Statement



Find all solutions to the equation below such that -180° \leq x \leq 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\epsilonI
 
Last edited:
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kylepetten said:

Homework Statement



Find all solutions to the equation below such that -180° \leq x \leq 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\epsilonI

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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