Solving trig equations

[TN17]

1. The problem statement, all variables and given/known data
I will ask two questions since they are both under the same lesson.

1) Determine exact solutions for each if domain is xe [0, 2Pi]
2tan^2x -12 = 0

2)Solve the following equations on interval xe [0, 2Pi]
sin x - sinxtanx = 0


2. Relevant equations
There aren't really any equations - just a lot of identities, lol.

3. The attempt at a solution
For 1)
2tan^2x = 12
tan^2x = 6
x = arctan (squareroot 6)
I had trouble with this one because (squareroot 6) isn't part of any special triangles, so I don't know how to get exact values from it.

For 2)
sin - sintan = 0
sin- sinsin/cos = 0
sincos - sin^2 / cos = 0
I had trouble with this one because there are two different identities instead of one. I didn't know how to continue from there.

Thank you.

[tiny-tim]

Hi TN17! :smile:

(have a pi: π and a square-root: √ and try using the X2 icon just above the Reply box :wink:)
1) Determine exact solutions for each if domain is xe [0, 2Pi]
2tan^2x -12 = 0
…
x = arctan (squareroot 6)
I had trouble with this one because (squareroot 6) isn't part of any special triangles, so I don't know how to get exact values from it.

I don't know tan-1√6 either :redface:

(Don't forget you need all the solutions in [0,2π])
2)Solve the following equations on interval xe [0, 2Pi]
sin x - sinxtanx = 0

just factor it! :wink:

[TN17]

[QUOTE=tiny-tim;2987316]Hi TN17! :smile:

(have a pi: π and a square-root: √ and try using the X2 icon just above the Reply box :wink:)


I don't know tan-1√6 either :redface:

(Don't forget you need all the solutions in [0,2π])


Yeah, I don't know how they got: pi/3, 2pi/3, 4pi/3 and 5pi/3

[TN17]

just factor it! :wink:[/QUOTE]
ok, well I got sin x -sinxtanx = 0
sin x- sinxsinx/cos = 0
sin x (1 - sinx/cos) = 0

...*Blank*.
I'm sorry, I'm pretty nervous for my test tomorrow. :S

[Mentallic]

Yeah, I don't know how they got: pi/3, 2pi/3, 4pi/3 and 5pi/3
Well those solutions suggest they were solving tan(x)=\pm\sqrt{3} not \sqrt{6}. Maybe you made a typo somewhere, like, it could be 2tan^2x-6=0 or (2tan(x))^2-12=4tan^2(x)-12=0


ok, well I got sin x -sinxtanx = 0
sin x- sinxsinx/cos = 0
sin x (1 - sinx/cos) = 0

...*Blank*.
I'm sorry, I'm pretty nervous for my test tomorrow. :S
Well if we have two products a and b such that ab=0, then what does this say about the solutions? Think back to when you would solve quadratics.

[TN17]

Well if we have two products a and b such that ab=0, then what does this say about the solutions? Think back to when you would solve quadratics.

Wow, mind block today.
Thanks, I solved it.
I feel pretty dumb for asking in the first place. :S

I didn't simplify tan. I did:
sinx - sinxtanx = 0
sinx(1-tanx) = 0
Case I:
sinx=0
x = 0, Pi, 2Pi

Case II:
1-tanx = 0
tanx = 1
x= Pi/4, 5Pi/4

[Mentallic]

Wow, mind block today.
Thanks, I solved it.
I feel pretty dumb for asking in the first place. :S

I didn't simplify tan. I did:
sinx - sinxtanx = 0
sinx(1-tanx) = 0
Case I:
sinx=0
x = 0, Pi, 2Pi

Case II:
1-tanx = 0
tanx = 1
x= Pi/4, 5Pi/4
Good, I was hoping you'll catch on and realize it's easier if you don't convert tan :smile:

By the way, with the first question, whether it was \sqrt{3} or \sqrt{6}, make sure that when you take the square root of both sides that you put in the \pm. You need to solve for both cases where tan(x)=\sqrt{3} and tan(x)=-\sqrt{3}. Remember that.