Solving Equations Using the Unit Circle

AI Thread Summary
The discussion focuses on solving the equation 2sin2x + sinx = 0 within the interval -180° ≤ x ≤ 90°. The initial factorization leads to sin x = 0 and sin x = -1/2 as potential solutions. The correct solutions for sin x = 0 are x = 0° and x = -180°, while for sin x = -1/2, the corresponding angle in the specified interval is -30°. Participants emphasize the importance of adhering to the given interval and correctly identifying solutions based on the unit circle. Overall, the conversation highlights the necessity of careful consideration of both the equation and the specified range in solving trigonometric equations.
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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
 
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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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