Help with identities

[umer007]

I'm sure its vry simple question but for sum reason I just cant solve it. 6cosx+5tanx=0 where x is equal or less than 2pie and equal or greater to -pie

 

3sinx-2cosx=0

then sinx/cosx=2/3, cosx can't =0

how can u move the sin's and cos's in one step like that and wt if it wz

3sinx+2cosx=0 then wt wd the eqn b?

 

Cud ne1 give me sum pointers on how to solve identities, i seem to be having sum problems with them. Ne help wud b much appreciated.

 

Thx.

[mezarashi]

I'm not sure what is happening here, because apparently this is not an identity. A trigonometric identity is one that is ALWAYS true, and denoted by the equal sign with three lines rather than just two.

 

The equation 6cosx + 5tanx = 0, does not even hold true for values of x between -pi < x < 2pi as you had originally stated. I think you are more or less trying to solve an equation for x is that right?

 

6cosx + 5tanx = 0

6cosx = -5tanx

6cosx = -5(sinx/cosx)

6cos^2(x) + 5sinx = 0 (use identity: sin^2 + cos^2 = 1, or cos^2 = 1 - sin^2)

6(1 - sin^2(x)) + 5sinx = 0

 

Let a = sin x

 

6 - 6a^2 + 5a = 0

6a^2 - 5a - 6 = 0

 

Solving using the quadratic eqation, you will get

a = 3/2, -2/3

 

Since sinx = a, and sinx is ALWAYS < 1, 3/2 is not a solution

thus:

 

sinx = a = -2/3

x = arcsin (-2/3)

 

If you try plugging this value, and ONLY this value into the original equation you will find that the equation is satisfied.

 

In anycase, a real trigonometric identity is one like: sin^2(x) + cos^2(x) = 1

Now you can replace x with ANY value you like, and the equation will ALWAYS be true.

 

Pointers about solving trigonometric identities. This process is actually termed "proving" them. You are NOT under any circumstance allowed to shift numbers across the identity (equal sign). For example in the example earlier: sin^2(x) + cos^2(x) = 1, I could not manipulate it into: sin^2(x) = 1 - cos^2(x) THIS IS ILLEGAL. Because you are trying to prove it is always true, by moving the cosine across, you are assuming that is already is, thus contradicting yourself and the proof. This question however was not an identity problem, rather just an exercising solving for x.

 

Please let me know if you need any more help.

[Dave]

Since sinx = a, and sinx is ALWAYS < 1, 3/2 is not a solution

 

Just nitpicking: [math]\sin(x) \leq 1[/math].

 

Rest of it is fine though

[Ducky Havok]

I'm sure its vry simple question but for sum reason I just cant solve it. 6cosx+5tanx=0 where x is equal or less than 2pie and equal or greater to -pie

 

3sinx-2cosx=0

then sinx/cosx=2/3' date=' cosx can't =0

how can u move the sin's and cos's in one step like that and wt if it wz

3sinx+2cosx=0 then wt wd the eqn b?

 

Cud ne1 give me sum pointers on how to solve identities, i seem to be having sum problems with them. Ne help wud b much appreciated.

 

Thx.[/quote']

 

Okay, this is just a personal pet peeve, but it would make your problem so much easier to read and work out if you actually spelt out all your words. For example, at first I thought by wt and wz you were talking about forms (which confused a simple minded person like myself), and it took a couple read throughs until I figured you meant what and was. Also, these look kind of like homework problems (judging from the 2nd problem), so they might do better in that thread.

 

Anyway, for the 2nd problem, they didn't do it in one step, they did it in 3, but just showed it as one. First they added [math]2cos{x}[/math], then they divided by [math]3[/math], then they divided by [math]cos{x}[/math]. Using an identity, you know that [math]\frac{sin{x}}{cos{x}}=tan{x}[/math], so you could just do [math]arctan{\frac{2}{3}}[/math] to get your answer.

 

If it was [math]3sin{x}+2cos{x}=0[/math], you would just subtract [math]2cos{x}[/math], divide by [math]3[/math], and divide by [math]cos{x}[/math], to get [math]\frac{sin{x}}{cos{x}}=-\frac{2}{3}[/math]. Then you can just replace [math]\frac{sin{x}}{cos{x}}[/math] with [math]tan{x}[/math].

 

I won't bother with the first since mezarashi already explained it quite well.

[Dave]

It would also help very much if you used LaTeX, although that's completely optional i suppose.

[bloodhound]

Just nitpicking: [math]\sin(x) \leq 1[/math].

 

Rest of it is fine though

just nitpicking.. only when the sin is constrained to R

[Dave]

I got pwnd.

 

Except we were already talking about sine/cosine in terms of the reals, so fnar