1. Nov 30, 2016

### Veronica_Oles

1. The problem statement, all variables and given/known data
Solve sin2x + sinx =0

2. Relevant equations

3. The attempt at a solution
I first factor and get
sinx(sinx + 1) = 0

sinx = 0
x = sin-1(0)
x= 0
x= π-0 = π

sinx = -1
x= sin-1(1)
x= π/2
x = π + π/2 = 3π/2

x = { 0, π, 3π/2 }

However when I look at the solution this is correct but they also added 2π and I'm not sure why this is?
It seems like for sinx = 0 they did x = 2π-0 = 2π why is this? I thought that if it's zero you obtain your other value through the sin quadrant.

2. Nov 30, 2016

### QuantumQuest

For $sinx = 0, x = n\pi$ because $sinx$ becomes zero for $n = 0, 1, 2, \cdots$
Can you conclude in an analogous way for $sinx + 1 = 0$?

3. Nov 30, 2016

### ehild

If X is a solution, or X+2kπ is also solution where k is integer. So the solutions of the problem are 0+2kπ; π+2nπ; 3π/2+2mπ,(k, m, n are integers).

Possibly, the problem asked the solutions in the closed interval [0, 2π]. In this case, both 0 and 2π have to be included.

4. Nov 30, 2016

### SammyS

Staff Emeritus
You wrote:
If they actually stated that in their solution, then that's in error and what you say after that is correct.

However, if it simply "seems" to you that they did this to get 2π, then I suspect that they did not use that reasoning, but merely used the fact that sin(x) has a period of 2π so that if x = 0 is a solution then x = 0 + 2π is also a solution.

5. Nov 30, 2016

### LCKurtz

You mean $x = \arcsin(-1)$, which doesn't give you $\frac \pi 2$. It gives you $x = -\frac \pi 2$. Your $\frac{3\pi} 2$ below is correct but your reasoning is incorrect.

6. Nov 30, 2016

### Veronica_Oles

x = sin-1(-1)
x = -π/2
x = π-(-π/2)
x = 3π/2

7. Nov 30, 2016

### LCKurtz

That gets $x=\frac \pi 2$ alright, but if I wanted an angle in $[0,2\pi]$, I would simply take $-\frac \pi 2 + 2\pi$, which would give $\frac {3\pi} 2$. I don't understand your thinking to calculate $\pi - (-\frac \pi 2)$. On what basis would you calculate that way?