- #1

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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?

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- Thread starter sportsguy3675
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- #1

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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?

- #2

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What exactly are you trying to do? Solve?

- #3

Zurtex

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Do what you said for the 2nd one and then let y = sin(x) and solve the quadratic equation.sportsguy3675 said:

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?

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

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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. :(

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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Really? I'm such a wimp the first thing with a trig equation is put everything back to sin and cos!Oh, my teacher told me sin and cos usually mess things up

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 cos

[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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Erm, you what? Let y = sin(x), you'll see you defintly don't get complex solutions.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. :(

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

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

- #8

Zurtex

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You have:sportsguy3675 said:

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

-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

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- #10

Zurtex

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To instill good methods that you will need for more advanced mathematics.sportsguy3675 said:

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