Question

Please check my answer (PreCal)

Answer 1

Solve csc x + 2 = 0 for 0 < x < 2pi.
I think it is 7pi/6 and 11pi/6

Answer 2

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

Answer 3

Thanks. Can you help with this one though, I am confused.
Solve 2 sin x + sqrt(3) < 0 for 0 x < 2.

Answer 4

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...

Answer 5

The answer choices are
\[\frac{ 4\pi }{3 } < x < \frac{ 5\pi }{ 3 }\]
\[\frac{ 2\pi }{3 } < x < \frac{ 4\pi }{ 3 }\]
\[\frac{ 7\pi }{6 } < x < \frac{ 11\pi }{ 6 }\]
\[\frac{ 5\pi }{6 } < x < \frac{ 7\pi }{ 6 }\]

Answer 6

so it would be a?

Answer 7

that would be my best guess... but its only a guess...

Answer 8

Thank you!