## Rosy95 Group Title Please check my answer (PreCal) one year ago one year ago

1. Rosy95

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

2. campbell_st

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

3. Rosy95

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

4. campbell_st

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...

5. Rosy95

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

6. Rosy95

so it would be a?

7. campbell_st

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

8. Rosy95

Thank you!