Solving Equations Using the Unit Circle

In summary, the solutions to the equation 2sin2x + sinx = 0 are x = 0° + 360°n and x = -180° + 360°n, where n is an integer. On the interval -180° <= x <= 90°, the only solutions are x = 0° and x = -180°. It is also important to note that the solution sinx = -1/2 was not included in the original problem description.
  • #1
kylepetten
25
0

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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  • #2
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?
 
  • #3
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
 
  • #4
kylepetten said:
im not using equations, i am getting my exact values from the unit circle
Yeah, Mark44! :smile:
 
  • #5
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.
 
  • #6
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.
 
  • #7
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.
 

Related to Solving Equations Using the Unit Circle

1. What is the Unit Circle?

The Unit Circle is a circle with a radius of 1 unit, centered at the origin on a Cartesian coordinate system. It is used in mathematics to understand and solve equations involving trigonometric functions.

2. Why is the Unit Circle important in solving equations?

The Unit Circle provides a visual representation of the relationships between trigonometric functions and the coordinates on the circle. This allows us to easily solve equations involving these functions and find exact solutions.

3. How do you use the Unit Circle to solve equations?

To use the Unit Circle to solve equations, we first need to understand the values of the trigonometric functions (sine, cosine, tangent, etc.) at each point on the circle. Then, we can graph the equation on the Unit Circle and use the coordinates of the points of intersection to find the solutions.

4. Can the Unit Circle be used to solve any equation involving trigonometric functions?

Yes, the Unit Circle can be used to solve any equation involving trigonometric functions. It is a powerful tool that allows us to find exact solutions to equations that would otherwise be difficult to solve.

5. Are there any tips for using the Unit Circle to solve equations?

One helpful tip is to memorize the coordinates of the points on the Unit Circle. This will make it easier to graph equations and find solutions. It is also important to understand the relationships between the trigonometric functions and their values on the Unit Circle.

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