## anonymous one year ago 3tanx3-10tanx=0 , in the domain 180<x<360, the measure of x is to the nearest degree?

1. anonymous

excuse me is it tan3x or tanx3

2. anonymous

tanx^3

3. anonymous

sorry i mean tanx^2

4. anonymous

so 2tanx^2 -10tanx

5. anonymous

omggg 3tanx^2 -10tanx

6. anonymous

@nincompoop

7. anonymous

@abb0t

8. jim_thompson5910

Is it $\Large 3\tan(x^2) - 10\tan(x)=0$ or Is it $\Large 3\left(\tan(x)\right)^2 - 10\tan(x)=0$ ??

9. anonymous

the second one

10. jim_thompson5910

you can let z = tan(x) that gives $\Large 3z^2 - 10z = 0$ $\Large z(3z - 10) = 0$ agreed?

11. anonymous

actually the x is the angle sign

12. anonymous

yep i get those steps dk what to di next

13. jim_thompson5910

so if z(3z-10) = 0, then either z = 0 or 3z-10 = 0 by the zero product property

14. jim_thompson5910

re-substitute in z = tan(x) z = 0 turns into tan(x) = 0 3z-10 = 0 turns into 3*tan(x) - 10 = 0

15. anonymous

jim we need to consider the xrange of x i also got here but the range got me confused

16. jim_thompson5910

true, that's a good point

17. anonymous

ok i get this whats next

18. jim_thompson5910

solve tan(x) = 0 and 3*tan(x) - 10 = 0 and make sure 180<x<360

19. jim_thompson5910

use the arctan function on a calculator to undo the tan function

20. anonymous

or we can put the range

21. anonymous

tan 0 i get 0 , for other i get 73.300

22. jim_thompson5910

is x = 0 in the interval 180<x<360 ?

23. jim_thompson5910

is x = 73.3 in the interval 180<x<360 ?

24. anonymous

umm

25. anonymous

yea that the part i am stuck on

26. jim_thompson5910

if it is, then you can keep the solution if not, then add/subtract 180 until you get in the interval

27. jim_thompson5910

being in the interval means that the number is between 180 and 360

28. anonymous

ok 73 is

29. jim_thompson5910

nope, 73 is not between 180 and 360

30. anonymous

no i mean when i add 180

31. jim_thompson5910

oh ok

32. anonymous

excuse since its periodic the range can be taken in 0<x<180 or 0<x<2pie

33. anonymous

k got the answer thank you

34. jim_thompson5910

no problem

35. anonymous

thanks jim

36. anonymous

u guy can stick around ? i got more

37. anonymous

oky woul love to in any way