# How to solve general equation questions?

• Mar 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
• Mar 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 $\displaystyle \sin{3x} \equiv 3\cos^2{x}\sin{x} - \sin^3{x}$.
• Mar 20th 2011, 02:32 AM
chris99191
i havnt been introduced to the sin3x identity so is there another way to do it
• Mar 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 $sin(3x)= 3sin(x)- 4sin^3(x)$.
• Mar 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 $A = \pi - B + 2n \pi$.
• Mar 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
• Mar 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
• Mar 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
• Mar 20th 2011, 03:15 PM
chris99191
i think in the form
if sin theta=sin a then theta=n.pi+(-1)^n.a
• Mar 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
• Mar 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.