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Trigonometric Equations

  • #1
I'm having trouble with two problems:
2tan(x) - 2cot(x) = -3
and
cos(x)^2 + sin(x) = 0
On the 2nd one, I can substitute 1-sin(x)^2 for cos(x)^2 right? I tried that, but it didn't work. And I have no clue what to do on the first one. Little help please?
 

Answers and Replies

  • #2
What exactly are you trying to do? Solve?
 
  • #3
Zurtex
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sportsguy3675 said:
I'm having trouble with two problems:
2tan(x) - 2cot(x) = -3
and
cos(x)^2 + sin(x) = 0
On the 2nd one, I can substitute 1-sin(x)^2 for cos(x)^2 right? I tried that, but it didn't work. And I have no clue what to do on the first one. Little help please?
Do what you said for the 2nd one and then let y = sin(x) and solve the quadratic equation.

For the 1st one, write all trig functions in terms of sin(x) and cos(x) usually is the way to go for these sort of things.
 
  • #4
Oh, my teacher told me sin and cos usually mess things up, but I guess that is the only option.

Edit:

I ended up with 2sin(x)^2 - 2cos(x)^2 + 3cox(x)sin(x) = 0 for the first one. That can't be right can it?

And using the identity on the 2nd one I get:

-sin(x)^2 + sin(x) +1 = 0

Then what, quad formula? But I got a negative under the radical. :(
 
Last edited:
  • #5
HallsofIvy
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Oh, my teacher told me sin and cos usually mess things up
Really? I'm such a wimp the first thing with a trig equation is put everything back to sin and cos!

In the first equation, 2tan(x) - 2cot(x) = -3, setting tan(x)= sin(x)/cos(x) and cos(x)/sin(x) you immediately get
[tex]2\frac{sin(x)}{cos(x}- 2\frac{cos(x)}{sin(x)}= -3[/tex]
Multiply through by sin(x) and cos(x) and you get
[tex]2sin^2(x)- 2cos^2(x)= 3[/tex]
Since cos2(x)= 1- sin2(x) the becomes
[tex]2sin^2(x)- 2+ 2sin^2(x)= 3[/tex]
so
[tex]4sin^2(x)= 5[/tex]
or
[tex]sin^2(x)= \frac{5}{4}[/tex]
That should be easy to solve.
 
  • #6
Zurtex
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sportsguy3675 said:
Oh, my teacher told me sin and cos usually mess things up, but I guess that is the only option.

Edit:

I ended up with 2sin(x)^2 - 2cos(x)^2 + 3cox(x)sin(x) = 0 for the first one. That can't be right can it?

And using the identity on the 2nd one I get:

-sin(x)^2 + sin(x) +1 = 0

Then what, quad formula? But I got a negative under the radical. :(
Erm, you what? Let y = sin(x), you'll see you defintly don't get complex solutions.

Eek, didn't realise you'd put Cot for the 2nd one, just multiply through my Tan(x) and apply simmilar princable to what we've just done.
 
  • #7
Um HallsofIvy how was the 3 seemingly unaffected by you multiplying through by sin and cos? And when I type in sin^-1(+/- root 5 / 2) I get domain error.... ?

What do you mean by let y=sin(x)?
 
  • #8
Zurtex
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sportsguy3675 said:
Um HallsofIvy how was the 3 seemingly unaffected by you multiplying through by sin and cos? And when I type in sin^-1(+/- root 5 / 2) I get domain error.... ?

What do you mean by let y=sin(x)?
You have:

-sin(x)^2 + sin(x) +1 = 0

Let y = sin(x):

-y^2 + y + 1 = 0

Solve y, then substitute sin(x) = y and then solve for x.

Do the same for the other one once you have multiplied through by tan(x)
 
  • #9
Yeah, I figured out the tan(x) thing. These were due today. I forget how I did the 2nd one because I didn't do it like you said. It worked some other way. What is the purpose of learning precal when the calculator already told me the answer :p
 
  • #10
Zurtex
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sportsguy3675 said:
Yeah, I figured out the tan(x) thing. These were due today. I forget how I did the 2nd one because I didn't do it like you said. It worked some other way. What is the purpose of learning precal when the calculator already told me the answer :p
To instill good methods that you will need for more advanced mathematics.
 

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