## anonymous 2 years ago tan theta + sec theta = 1 between 0 and 2 pi

1. anonymous

can theta=0 ? Like is is$0< \theta <\Pi$ Or $0 \leq \theta \leq \Pi$ ?

2. anonymous

0<θ<2Π has to be between this interval

3. anonymous

use substitution y=tan(theta/2) |dw:1383969602615:dw|

4. anonymous

0≤θ<2Π sorry correction, it has to be between this interval

5. anonymous

with a little bit of work you will find the solution

6. anonymous

@Mertsj what about if they have a log like for example using the graph solve the system y=-2x+3 and y=logx+1 im confused cause of the log

7. anonymous

@RED123url I am not sure what you did there...

8. anonymous

well y=tan(theta/2) sin(thata)= 2y/(y^2+1) cos(theta)=(1-y^2)/(y^2+1) you know tan=sin/cos sec=1/cos plug things in and you will get an equation of y then go back y=tan)theta/2) and use the inverse

9. anonymous

The answer is 0 but I used trial and error

10. anonymous

@RED123url so just to confirm what swissgirl got, did you get the same answer?

11. anonymous

I graphed it to confirm my answer and its def right

12. anonymous

it should be 2pi

13. anonymous

Well it is 0 and 2 pi But if you noticed the interval is $0 \leq \theta < 2\pi$

14. anonymous

@RED123url well 2pi is not one the answers but 0 is, also according to the interval 2pi cannot be the answer..

15. anonymous

the real answer for these kind of problem involves n: the answer is 2npi. n can be 0 or any number even a billion

16. anonymous

@RED123url so would you agree that in this case the answer is 0?

17. anonymous

if 2pi is excluded of course

18. anonymous

well then. I am going with 0. Thank you both for your help!