Solve for x with a trigonometric 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) ##.
 
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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).
 
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