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
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 $\displaystyle \displaystyle \sin{3x} \equiv 3\cos^2{x}\sin{x} - \sin^3{x}$.
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 $\displaystyle sin(3x)= 3sin(x)- 4sin^3(x)$.
$\displaystyle \sin(3x) + \sin(2x) = 0 \Rightarrow \sin(3x) = -\sin(2x) \Rightarrow \sin(3x) = \sin(-2x)$.
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 $\displaystyle A = \pi - B + 2n \pi$.
$\displaystyle x = 2n \pi$ 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