How to solve general equation questions?

Printable View

March 19th 2011, 11:31 PM
chris99191

How to solve general equation questions?

Find the general solution to sin3x+sin2x=0

I cant figure out where to start and how to solve it. I think maybe factorising to sin(3x+2x) but im not sure
March 19th 2011, 11:36 PM
Prove It

First of all $\displaystyle \sin{3x} + \sin{2x} \not \equiv \sin{(3x + 2x)}$ .

You will need to use the identities $\displaystyle \sin{2x} \equiv 2\sin{x}\cos{x}$ and $\displaystyle \sin{3x} \equiv 3\cos^2{x}\sin{x} - \sin^3{x}$ .
March 20th 2011, 02:32 AM
chris99191

i havnt been introduced to the sin3x identity so is there another way to do it
March 20th 2011, 02:57 AM
HallsofIvy

Because you have sin(3x), you are going to have to, in some way, use that identity.

Do you have sin(2x)= 2sin(x)cos(x) and $cos(2x)= cos^2(x)- sin^2(x)$ ? Do you have sin(x+ y)= sin(x)cos(y)+ cos(x)sin(y)? If so then you know that $sin(3x)= sin(2x+ x)= sin(2x)cos(x)+ cos(2x)sin(x)= (2sin(x)cos(x))cos(x)+ (cos^2(x)- sin^2(x))sin(x)$ $= 2sin(x)cos^2(x)+ sin(x)cos^2(x)- sin^3(x)= 3sin(x)cos^2(x)- sin^3(x)$ . And, of course, $cos^2(x)= 1- sin^2(x)$ so that says that $sin(3x)= 3sin(x)- 4sin^3(x)$ .
March 20th 2011, 04:02 AM
mr fantastic

Quote:

Originally Posted by chris99191

Find the general solution to sin3x+sin2x=0

I cant figure out where to start and how to solve it. I think maybe factorising to sin(3x+2x) but im not sure

$\sin(3x) + \sin(2x) = 0 \Rightarrow \sin(3x) = -\sin(2x) \Rightarrow \sin(3x) = \sin(-2x)$ .

Case 1: $3x = -2x + 2n \pi \Rightarrow x = ....$

Case 2: $3x = \pi - (-2x) + 2n \pi \Rightarrow x = ....$

In both cases n is an integer.

NB: If $\sin(A) = \sin(B)$ then either $A = B + 2n \pi$ or $A = \pi - B + 2n \pi$ .
March 20th 2011, 02:14 PM
chris99191

Thankyou for your reply but HallsofIvy your end solution is not the general solution which is what is required in this exercise im doing and mr fantastic its the same with you
March 20th 2011, 02:18 PM
chris99191

Thankyou both of you for your help but your end answers are not in the general solution form, which is the requirement for this exercise im doing
March 20th 2011, 03:06 PM
topsquark

Quote:

Originally Posted by chris99191

Thankyou for your reply but HallsofIvy your end solution is not the general solution which is what is required in this exercise im doing and mr fantastic its the same with you

What "general" solution do you need? They are as general an answer as any I can think of.

-Dan
March 20th 2011, 03:15 PM
chris99191

i think in the form
if sin theta=sin a then theta=n.pi+(-1)^n.a
March 20th 2011, 04:40 PM
topsquark

Quote:

Originally Posted by chris99191

i think in the form
if sin theta=sin a then theta=n.pi+(-1)^n.a

$x = 2n \pi$ or $x = 2n \pi + \pi$
(with apologies to mr fantastic.)

There is no more general solution to sin(3x) + sin(2x) = 0 than this one.

The only place in trigonometry I have ever run into $(-1)^n$ is when we are talking about the sine or cosine of an angle, not what the angle itself is. For example
$cos(n \pi) = (-1)^n$
where n is an integer, so I don't see how such an expression could get into the form for an angle.

-Dan
March 20th 2011, 06:46 PM
mr fantastic

Quote:

Originally Posted by chris99191

Thankyou for your reply but HallsofIvy your end solution is not the general solution which is what is required in this exercise im doing and mr fantastic its the same with you

The solutions you get from my reply (you need to fill in the gaps) are the general solutions. Substitute values of n and see for yourself.