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Please check my answer (PreCal)

Mathematics
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Solve csc x + 2 = 0 for 0 < x < 2pi. I think it is 7pi/6 and 11pi/6
so you have csc(x) = -2 so sin(x) = -1/2 this is 3rd and 4th quadrants so pi + pi/6 = 7pi/6 (3rd quadrant) 2pi - pi/6 = 11pi/6 (4th Quadrant) so you solution looks great... well done
Thanks. Can you help with this one though, I am confused. Solve 2 sin x + sqrt(3) < 0 for 0 x < 2.

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Other answers:

ok... so you will start with \[2\sin(x) < -\sqrt{3} \] so \[\sin(x) < - \frac{\sqrt{3}}{2}\] so aren't there 2 inequaitites to this in the 3rd quadrant \[\pi < x < \frac{4\pi}{3}\] and the 4th quadrant \[\frac{5\pi}{3} < x < 2\pi\] tough question, I not really confident on the solutions...
The answer choices are \[\frac{ 4\pi }{3 } < x < \frac{ 5\pi }{ 3 }\] \[\frac{ 2\pi }{3 } < x < \frac{ 4\pi }{ 3 }\] \[\frac{ 7\pi }{6 } < x < \frac{ 11\pi }{ 6 }\] \[\frac{ 5\pi }{6 } < x < \frac{ 7\pi }{ 6 }\]
so it would be a?
that would be my best guess... but its only a guess...
Thank you!

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