# Thread: How to solve general equation questions?

1. ## 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

2. 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 $\displaystyle \displaystyle \sin{3x} \equiv 3\cos^2{x}\sin{x} - \sin^3{x}$.

3. i havnt been introduced to the sin3x identity so is there another way to do it

4. 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 $\displaystyle sin(3x)= 3sin(x)- 4sin^3(x)$.

5. 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
$\displaystyle \sin(3x) + \sin(2x) = 0 \Rightarrow \sin(3x) = -\sin(2x) \Rightarrow \sin(3x) = \sin(-2x)$.

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 $\displaystyle A = \pi - B + 2n \pi$.

6. 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

7. 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

8. 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

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

10. Originally Posted by chris99191
i think in the form
if sin theta=sin a then theta=n.pi+(-1)^n.a
$\displaystyle x = 2n \pi$ 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

11. 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.