Question

if you divide 2sin2x by 2 do you get sinx or sin2x?

Answer 1

Like this?
\[\large \frac{2\sin(2x)}{2}\]

You only perform the division once, meaning ~ on one of the factors in our multiplication on top.\[\Large =\sin(2x)\]

Think about this to reinforce that idea,\[\Large \frac{2\cdot2}{2} =2\]We don't divide each of the 2's in the numerator by the 2 in the denominator.

Answer 2

yes. thank you

Answer 3

Wait. can you help me finish the problem? Im confused

Answer 4

What is the rest of the problem? :o

Answer 5

okay it was 2sin2x+sqrt3=0
I subtracted the sqrt of 3 and got 2sin2x=-sqrt3
then I divided by 2 and got sin2x=-sqrt3/2
Now Im not sure what to do next

Answer 6

I have to find the 4 solutions between 0 and 2pi

Answer 7

What is the rest of the problem? :o

Answer 8

Thats it. It said find the 4 solutions between 0 and 2pi

Answer 9

AHHHHHHHHHHH my openstudy crashed again :( lost all that typing grrr

Answer 10

\[\Large \sin(2x)=-\frac{\sqrt3}{2}\]This value corresponds to a special angle on the unit circle.
This next step might be a little bit confusing since our angle in this case is not \(\large \theta\) but instead is \(\large 2x\).

Do you remember which places on the unit circle will give us this value when we take the sine?

Answer 11

Take the sin of -sqrt3/2?

Answer 12

Do you just ignore the 2 before the x?

Answer 13

Let's ignore it for now,
\[\large \sin u=-\frac{\sqrt3}{2}\]
If you had this expression, what special angles do we get?

Answer 14

4pi/3 and 5pi/3

Answer 15

Good, we would say our `angle` \(\large u=4\pi/3\) and \(\large u=5\pi/3\).

But our angle in this problem is the thing inside of the sine function.
So we end up with,\[\large 2x=4\pi/3 \qquad\text{and}\qquad 2x=5\pi/3\]

Answer 16

Our next step would be to solve for x.

Confused by the 2x thing at all? :O

Answer 17

yes.. Im lost with that 2 there

Answer 18

so wait. are 4pi/3 and 5pi/3 solutions at all?

Answer 19

\[\large \sin(Angle)=-\frac{\sqrt3}{2}\]And you determined that,\[\Large Angle=4\pi/3 \qquad\text{and}\qquad Angle=5\pi/3\]

See how our angle in this problem is 2x?
\[\Large \sin(2x)=-\frac{\sqrt3}{2}\]Giving us,
\[\Large 2x=4\pi/3 \qquad\text{and}\qquad 2x=5\pi/3\]

Answer 20

No they are not solutions.
We haven't determined solutions yet :o