Solve for x with a trigonometric function

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Homework Help Overview

The discussion revolves around solving the equation sin(3x) = -1/2, focusing on finding all possible values of x. The subject area is trigonometric functions and their properties, particularly in relation to angles in different quadrants.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore solutions for angles in both the 4th and 3rd quadrants. There is a suggestion to use a more systematic approach to ensure all solutions are found, including the consideration of negative integer values for n in the general solution.

Discussion Status

The discussion is ongoing, with participants providing feedback on the original poster's solutions and suggesting the need for completeness in the solution set. Some participants express uncertainty about specific values while others acknowledge the need to include negative solutions.

Contextual Notes

Participants note that the original poster's solutions may be incomplete and emphasize the importance of considering all quadrants and the periodic nature of the sine function.

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Homework Statement
Solve for x.
Relevant Equations
sin(3x)= -1/2
Homework Statement: Solve for x.
Homework Equations: sin(3x)= -1/2

sin(3x) = -1/2

3x = sin-1(-1/2)

3x = -π/6

x = -π/18

x = -π/18 + 2π/3 = 11π/18

11π/18 + 2π/3 = 23π/18

11π/18 + (2π(4))/3 = 35π/18

The solutions I obtained were 23π/18 and 35π/18. Are these correct? I'm not entirely sure if I did this problem correctly. Thank you.
 
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What about a solution or solutions where the angle 3x is in the 3rd quadrant? (You found the solution when 3x is in the 4th quadrant). ## \\ ## Your solutions are incomplete, even for the 4th quadrant. You need to let ##3x=\theta+n (2 \pi) ##, where ## n ## is any integer. You found ## \theta ## for the 4th quadrant, etc., but it helps to be more systematic. Otherwise, you don't get the complete set.
 
Charles Link said:
What about a solution or solutions where the angle 3x is in the 3rd quadrant? (You found the solution when 3x is in the 4th quadrant). ## \\ ## Your solutions are incomplete, even for the 4th quadrant. You need to let ##3x=\theta+n (2 \pi) ##, where ## n ## is any integer. You found ## \theta ## for the 4th quadrant, etc., but it helps to be more systematic. Otherwise, you don't get the complete set.
Yes I can see where I went wrong. I got the solutions 7π/18, 11π/18, 19π/18, 23π/18, 31π/18, 35π/18 after looking it over. Thank you.
 
Still incomplete. And please show your work. And I don't think I agree with the ## \frac{7 \pi}{18} ##. Edit: My mistake. OP has it correct. As @PAllen mentions below though, the OP needs to include the negative x values that come from negative integer ##n ## in ## 3x=\theta+n(2 \pi) ##.
 
Last edited:
Charles Link said:
And I don't think I agree with the ## \frac{7 \pi}{18} ##.
Looks fine to me. Also, his last set of solutions looks fine as the beginning of an obvious pattern that is complete (except for negative x values).
 
Last edited:
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PAllen said:
Looks fine to me. Also, his last set of solutions looks fine as the beginning of an obvious pattern that is complete (except for negative x values).
Sorry. My mistake. See the Edit above.
 

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