How Do You Solve 26cot²Θ = cosec²Θ for -π ≤ Θ ≤ π?

[lemon]

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

Homework 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?

 

[Mark44]

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

Homework 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

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

Guess that was a wrong start. Any suggestions please?

 

[lemon]

26cos²Thetasin²Theta=sin²Theta
= 26cos²Theta

really?

 

[Mark44]

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

 

[lemon]

so cot²Theta=(cos²Theta/sin²Theta)?

 

[Mark44]

Yes, and csc²(theta) = 1/sin²(theta)

 

[lemon]

How about this?
26cot²x=1+cot²x
25cot²x-1=0
25/tan²x-1=0
25-tan²x=tan²x
2tan²x-25=0

can I go with that?

 

[Mark44]

How about this?
26cot²x=1+cot²x
25cot²x-1=0

25/tan²x-1=0

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?

25-tan²x=tan²x
2tan²x-25=0

can I go with that?

No.

 

[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?

 

[lemon]

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

 

[Mark44]

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?

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.

 

[Mark44]

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

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).

 

[lemon]

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

 

[lemon]

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

 

[lemon]

tanx=5
Thanks Mark44

 

[lemon]

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

 

[Mark44]

tanx=5
Thanks Mark44

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

 

[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.)

 

[Mark44]

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.

 

[lemon]

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

 

[Mark44]

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