Solving Quadratic Trig Equation

[Veronica_Oles]

Homework Statement

Solve sin²x + sinx =0

Homework Equations

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.

 

[QuantumQuest]

Homework Statement

Solve sin2x + sinx =0

Homework Equations

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.

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

 

[ehild]

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.

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.

 

[SammyS]

You wrote:

It seems like for sin x = 0 they did x = 2π-0 = 2π ...

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.

 

[LCKurtz]

Homework Statement

Solve sin²x + sinx =0

Homework Equations

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)

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.

x= π/2
x = π + π/2 = 3π/2

x = { 0, π, 3π/2 }

 

[Veronica_Oles]

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.

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

 

[LCKurtz]

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

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?