Question

Find all solutions to the equation. sin2x + sin x = 0

Answer 1

sin2x = -sinx
hint: sin2x = 2sinxcosx

Answer 2

Have you tried using the hint, then setting the two equations equal to one another?

Answer 3

No I dont understand it at all can you please explain it?

Answer 4

@hieuvo suggested using the identity sin (2x) = 2 sin x cos x

So that would make your original equation

2 cos x sin x + sin x = 0,
or 2 cos x sin x = -sin x

Answer 5

Okay so how can I solve the equation?

Answer 6

eliminate sinx

Answer 7

Are you sure the question isn't: sin^2(x)+sin(x)=0?

Answer 8

That would change things a bit.

Answer 9

Yeah, and it is perfectly easily solvable too.

Answer 10

the problem says sin2x + sin x = 0

Answer 11

I mean sin^2x+ sin x =0

Answer 12

There are several ways you can do this. You can solve it like you would with the quadratic formula, or complete the square. You can substitute in something else for sinx so that it is more comfortable like the letter u.

u^2+u=0 if you say u=sinx, then plug in sin(x) in the end to get your answers if that makes sense.

Answer 13

Im sorry can you explain that a little more plaese?

Answer 14

@Kainui suggested letting u = sin(x)

Then your original problem becomes

u^2 + u = 0

Answer 15

Just approach this like you would any old quadratic formula solving for x:

ax^2+bx+c=0

except instead of x's you have sinx's that you're solving for.

Answer 16

Oh okay i think I get it now

Answer 17

Another way to think about this is to factor it, and use the zero product property.\[2 \cos x \sin x + \sin x = 0=\sin x(2\cos x+1) \implies \sin x =0~or~\cos x=-\frac{1}{2}\]This give us the solutions\[x \in \left\{ k \pi, \pm ( \frac{2 \pi}{3}+k \pi) \right\}~~ \forall k \in Z\]

Answer 18

what is k?

Answer 19

sinx (sinx+1) = 0
sinx = 0
sinx = -1
solve for x

Answer 20

Think of what angles, x, would cause sin(x) to be 0, or -1

Answer 21

Would it be pi/2?

Answer 22

pi/2 is 90 degrees. That is when sin(x) = 1. What angle causes sin(x) to be 0? And what angle causes sin(x) to be -1 ?

Answer 23

hint: unit circle
:3

Answer 24

180 degrees

Answer 25

Right. sin(x) = 0 at 0 degrees and 180 degrees on the circle

What about when sin(x) = 1?

Answer 26

Remember \(\sin(x)\) is a periodic function with period \(2\pi\).
This means if \(a\) is a solution, then \(\forall n \in \mathbb{N}\quad a+2\pi n\) is a solution.

Answer 27

In degrees, the period is \(360^\circ \).

Answer 28

\[ \forall n \in \mathbb{Z} \]

Answer 29

I am confused now..can you please explain?

Answer 30

First of all, did you get the normal solutions?

Answer 31

were they 0, 180, and 90?

Answer 32

Yes.

Answer 33

My point is 0+360 is a solution too. 0+2*360 is also a solution. 0-360 is a solution.

Does that make sense?

Answer 34

okay so all of these are solutions?

Answer 35

Yes, because if you rotate 360 degrees, you're back where you started.

Answer 36

ohhhhhh okay

Answer 37

So your solutions, plus any number of full rotations is a solution.

Answer 38

So what you do is say that \(n\) is any integer... such as 0, 1, -1, 2, -2, etc.

Answer 39

Then ALL solutions would be: \[
0+360^\circ n, 90^\circ +360^\circ n, 180^\circ + 360^\circ n
\]

Answer 40

So originally you determined

sin(x) = 0
sin(x) = -1

@wio represented all of the solutions for when sin(x) = 0.
For what values does sin(x) = -1 ?

Answer 41

It should have been sin(x) = -1, instead of sin(x) = 1 earlier

Answer 42

oooh, whoops, looks like 90 isn't a solution.

Answer 43

I think I mistyped it in an earlier post. Can't edit them without deleting it though.

Answer 44

I really wish I could edit posts...
I make little mistakes all the time.

Answer 45

So the solutions would be 0, 180, 0 + 360?

Answer 46

If we restrict our domain to one revolution around the circle, then x = 0 and x = 180 degrees are two solutions.

The other solutions we would obtain by setting

sin(x) = -1

For what value(s) of x does sin(x) = -1?

Answer 47

is it 180?

Answer 48

Nope... think of a circle