# Homework Help: Solve cot/cosec

1. Jan 31, 2010

### lemon

1. Solve 26cot2Theta=cosec2Theta for -π≤0≤π

2. Relevant equations

3. Well. I tried to solve this and ended up with 26cos2Theta=0. That can't be right.
I started with 26(cos2Theta/sin2)Theta=1/sin2Theta

Guess that was a wrong start. Any suggestions please?

2. Jan 31, 2010

### Staff: Mentor

Multiply both sides by sin2(theta). When you multiply the right side by sin2(theta) you don't get 0.

3. Jan 31, 2010

### lemon

26cos2Thetasin2Theta=sin2Theta
= 26cos2Theta

really?

4. Jan 31, 2010

### Staff: Mentor

No, that's not it. cot(theta) = cos(theta)/sin(theta), right? And csc(theta) = 1/sin(theta). Those are the identities to use.

5. Jan 31, 2010

### lemon

so cot2Theta=(cos2Theta/sin2Theta)?

6. Jan 31, 2010

### Staff: Mentor

Yes, and csc2(theta) = 1/sin2(theta)

7. Feb 3, 2010

### lemon

26cot2x=1+cot2x
25cot2x-1=0
25/tan2x-1=0
25-tan2x=tan2x
2tan2x-25=0

can I go with that?

8. Feb 3, 2010

### Staff: Mentor

In the next step after the line above, you should add 1 to both sides. After that, multiply both sides by tan2x.

The line below is wrong, which makes the line after it wrong, too. How did you get from the line above to this one?
No.

9. Feb 3, 2010

### lemon

I got to the next line by multiplying out tanx2 from the division.
Why would I add 1 to both sides. It will get rid of the -1 and put 1 to the right of the equation?

10. Feb 3, 2010

### lemon

That would give:
26cot2x=1+cot2x
25cot2x-1=0
(25/tan2x)-1=0
(25/tan2x)=1
right?

11. Feb 3, 2010

### Staff: Mentor

25/tan2(x) - 1 = 0
==> 25/tan2(x) = 1
==> 25 = tan2(x)
So tan(x) = ?

Doing it the way you did it,
25/tan2(x) - 1 = 0
==> 25 - tan2(x) = 0 Note that 0*tan2(x) is not tan2(x).
Now you want to add tan2(x) to both sides, which gets you to my last line above.

12. Feb 3, 2010

### Staff: Mentor

This would be confusing to a casual reader. All of the things you have as cot2x and tan2x are really cot(x) and tan2(x).

13. Feb 3, 2010

### lemon

ahh wait - sorry
I'm having issues with the editing software here.

14. Feb 3, 2010

### lemon

That would give:
26cot2x=1+cot2x
25cot2x-1=0
(25/tan2x)-1=0
(25/tan2x)=1
right?

15. Feb 3, 2010

### lemon

tanx=5
Thanks Mark44

16. Feb 3, 2010

### lemon

so my solutions should be at 1.37 (2d.p.)
and -π+1.37 = -1.77 (2d.p.)

17. Feb 3, 2010

### Staff: Mentor

That's one solution. What's the other one?

18. Feb 3, 2010

### lemon

err.. don't follow

I gave two solutions
solutions should be at 1.37 (2d.p.)
and -π+1.37 = -1.77 (2d.p.)

19. Feb 3, 2010

### Staff: Mentor

The equation tan2(x) = 25 has two solutions in terms of tan(x), and you got one of them, which led to two values of x.

There are two more values of x in the interval -pi <= x <= pi, meaning there are four solutions all together.

20. Feb 4, 2010

### lemon

ahh! Tanx=+or -5
So the other two solutions are at -1.37 and 1.77 (2d.p.) :tongue2:

21. Feb 4, 2010

### Staff: Mentor

IMO, little mistakes and omissions and such are doing you in. This is not a very complicated problem, yet it took 20+ posts to get to the solution.

Here is all you needed to do:

Solve 26 cot2(t) = csc2(t), -pi <= t <= pi
==> 26cos2(t)/sin2(t) = 1/sin2(t)
==> 26cos2(t) = 1
==> cos2(t) = 1/26
==>cos(t) = +/-sqrt(1/26)
==> t $\approx$ +/-1.3734, t $\approx$ +/ 1.7682