Question

Solve sin(x)(sinx + 1)=0.

Answer 1

zero product property says that \(\sin x=0 \) or \(\sin x=-1 \) ....

Answer 2

Where are you confused on what to do first?

Answer 3

I'm not really sure... I don't know what to do first

Answer 4

The advice that @mukushla provided is exactly how to go about doing this problem. Do you understand what he said?

Answer 5

Yes, I understand the zero product property. So to solve that, what would I do? Would I have to do sin^-1(0) and sin^-1(-1) to find x?

Answer 6

Oh no. So the first solution we have is
\[\huge \sin(x)=0\]
To solve for x we take the arcsin of both sides. Have you learned that yet?

Answer 7

Yeah I learned it, but I don't really understand it

Answer 8

s\[sinx=0 \implies x=\pm n \Pi \] where n is any integer

Answer 9

If not, we should know for what x makes sin equal to zero. Hint: consult the unit circle. :D

Answer 10

@MathFreak106 https://docs.google.com/open?id=0B98cXMBCs13BZThlZDAyMGUtZGIxMi00NjVhLWFjODktMGU4NjQ0ZDQzMWU5

This is a nice formula sheet i made a couple years ago.

Answer 11

\[sinx=-1 \implies x=3\pi/2.....\]

Answer 12

Lets look at arcsin for a moment.
if sin(x)=y THEN, arcsin(y)=x Where x and y aren't coords but any number or function.

Answer 13

Ok so arcsin(0)=x

Answer 14

RIGHT! and the arcsin of a function will find you an angle. Do you see why?

Answer 15

Kind of. How do you solve arcsin?

Answer 16

Do you have a unit circle around you?

Answer 17

Yep, from your worksheet :)

Answer 18

great! Lets look at this example
\[\huge \sin(x) = {1 \over 2}\]
Can you find the two angles where that happens?

Answer 19

Maybe it's better if i write it like:
\[\huge \sin(\theta)= {1 \over 2}\]

Answer 20

Umm 5pi/6?

Answer 21

Ya, thats one of them. There's one more.

Answer 22

I don't know what the other one is

Answer 23

check out \[\pi \over 6\]

Answer 24

Remember Theta can be in radians or degrees too. So looking at the unit circle again it could be 30 degrees or 150 degrees.

So if sin(x)=0 that means it could be pi or 0, OR 180 degrees or 0 degrees.

But if you look at the other side of my sheet you will see that arcsin has a range from
\[\huge [-\pi/2, \pi/2]\] which means that it can ONLY be ONE of the choices we have.