## Babynini one year ago Find all solutions for equation, find solutions in interval [0,2pi) sq3 tan 3(theta)+1=0

1. Babynini

@peachpi the victory was short lived lol

2. Babynini

3(theta)=arctan(-sq3/3) ?

3. anonymous

This is the original? $\sqrt{3\tan 3\theta+1}=0$

4. Babynini

hm, no. Let me write it out for you :) $\sqrt{3}\tan3(\theta)+1=0$

5. anonymous

got you

6. Babynini

$\tan3(\theta)=\frac{ -\sqrt{3} }{ 3 }$

7. Babynini

right?

8. anonymous

yes

9. Babynini

So then after quite a bit of simplifying theta = -0.26 ?

10. Babynini

oh wait -0.17

11. anonymous

That one's on the unit circle, so they might be looking for an exact answer. -π/18, which is about -0.17, so you're right

12. Babynini

pi/18 is on the unit circle? o.o

13. anonymous

Well pi/6 is.

14. anonymous

You had 3Θ = arctan (-1/√3) 3Θ = -π/6 Θ = -π/18

15. Babynini

Why does 3(theta) = -pi/6 ?

16. Babynini

sorry, these are probably really elementary questions but I want to make sure and understand.

17. anonymous

because arctan has a domain from -pi/2 to pi/2 (because tan positive in the 1st quadrant, negative in the 4th). The angle in that domain with a tan of -1/√3 is -π/6

18. Babynini

hm..k.

19. anonymous

does that make sense?

20. Babynini

Yeah I just hadn't made the connection so now i've got to think it through a bit :P

21. Babynini

So now to find it in the right quadrants we add pi?

22. anonymous

ok so this one's a little trickier since the tan is negative. The restriction on the equation is 0 to 2pi, so we actually need to add 2pi to -pi/6 to get the 1st solution for 3pi

23. anonymous

We'll have 2 solutions for 3pi between 0 and 2pi: 11π/6 in the 4th quadrant (-pi/6 + 2π) 5π/6 in the 2nd quadrant (-π/6 + π) |dw:1434608981061:dw|

24. anonymous

|dw:1434609190233:dw|

25. anonymous

Dividing each of those by 3 gives Θ = 11π/18 and Θ = 5π/18

26. anonymous

Make sense so far?

27. Babynini

and those aren't negative.

28. anonymous

right. we don't want the negative solutions

29. anonymous

You need to find 3Θ for 0<Θ<6π, so add 2pi to both 5π/6 and 11π/6 twice to get the other 4 solutions for 3Θ

30. Babynini

so it would be writen as 5pi/18 + 2kpi ?

31. anonymous

no just 5pi/18. because they gave a specific interval, you don't use 2kpi

32. Babynini

ou.

33. Babynini

but we're missing two more quadrants still, right?

34. anonymous

there are 4 more solutions. You need to find 3Θ for 0<Θ<6π, so add 2pi to both 5π/6 and 11π/6 twice to get the other 4 solutions for 3Θ

35. Babynini

....17pi/18 ?

36. anonymous

yes. that's a solution |dw:1434609720505:dw|

37. Babynini

23pi/18 ?

38. anonymous

yes

39. Babynini

so our four answers are 5pi/18, 11pi/18. 17pi/18, 23pi/18

40. anonymous

you've got 2 more to go

41. Babynini

29pi/18?

42. Babynini

35pi/18

43. anonymous

you got it :)

44. Babynini

yaya thaanks.

45. anonymous

no problem