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

Printable View

- Mar 19th 2011, 11:31 PMchris99191How 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 PMProve It
First of all $\displaystyle \displaystyle \sin{3x} + \sin{2x} \not \equiv \sin{(3x + 2x)}$.

You will need to use the identities $\displaystyle \displaystyle \sin{2x} \equiv 2\sin{x}\cos{x}$ and $\displaystyle \displaystyle \sin{3x} \equiv 3\cos^2{x}\sin{x} - \sin^3{x}$. - Mar 20th 2011, 02:32 AMchris99191
i havnt been introduced to the sin3x identity so is there another way to do it

- Mar 20th 2011, 02:57 AMHallsofIvy
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 $\displaystyle cos(2x)= cos^2(x)- sin^2(x)$? Do you have sin(x+ y)= sin(x)cos(y)+ cos(x)sin(y)? If so then you know that $\displaystyle 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)$$\displaystyle = 2sin(x)cos^2(x)+ sin(x)cos^2(x)- sin^3(x)= 3sin(x)cos^2(x)- sin^3(x)$. And, of course, $\displaystyle cos^2(x)= 1- sin^2(x)$ so that says that $\displaystyle sin(3x)= 3sin(x)- 4sin^3(x)$. - Mar 20th 2011, 04:02 AMmr fantastic
$\displaystyle \sin(3x) + \sin(2x) = 0 \Rightarrow \sin(3x) = -\sin(2x) \Rightarrow \sin(3x) = \sin(-2x)$.

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

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

In both cases n is an integer.

**NB:**If $\displaystyle \sin(A) = \sin(B)$ then either $\displaystyle A = B + 2n \pi$ or $\displaystyle A = \pi - B + 2n \pi$. - Mar 20th 2011, 02:14 PMchris99191
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 PMchris99191
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 PMtopsquark
- Mar 20th 2011, 03:15 PMchris99191
i think in the form

if sin theta=sin a then theta=n.pi+(-1)^n.a - Mar 20th 2011, 04:40 PMtopsquark
$\displaystyle x = 2n \pi$ or $\displaystyle x = 2n \pi + \pi$

(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 $\displaystyle (-1)^n$ is when we are talking about the sine or cosine of an angle, not what the angle itself is. For example

$\displaystyle cos(n \pi) = (-1)^n$

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 PMmr fantastic