Question

What values for q (0 < q < 2p) satisfy the equation?

2sqrt(2)sin q + 2 = 0

Answer 1

\(\bf \large 2\sqrt{2}sin(q)+2=0\ \ ;\ \ 0 < q < 2\pi\)

Answer 2

Can someone teach me how to solve this? I would like to be able to solve it on my own bu the text book is just not making sense today.

Answer 3

do the regular algebra by isolating sin(q) on one side of the equal sign....

Answer 4

\(\bf \large {2\sqrt{2}sin(q)+2=0; 0 < q < 2\pi\\
sin(q) = \frac{2}{2\sqrt{2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}}\)

Answer 5

that should be negative on the right side... ie, \(\frac{-2}{2\sqrt2} \)

Answer 6

wait! so the first step is to divide by \[2\sqrt{2}\] to move it to the right??

Answer 7

actually, yes, should be -2 :/

Answer 8

no... move the lone "2" first by subtractig...

Answer 9

ok i see that which is where the nominator comes from.

Answer 10

\(\bf \large 2\sqrt{2}sin(q)+2=0; 0 < q < 2\pi\\
\large sin(q) = -\frac{2}{2\sqrt{2}} = -\frac{1}{\sqrt{2}} = -\frac{\sqrt{2}}{2}\)

Answer 11

yep...

Answer 12

how did the \[\sqrt{2}\] get on top of the fraction?

Answer 13

then for that, just check your Unit Circle for the angles between 0 and \(\bf \large \pi\) that have that value with a NEGATIVE sign

Answer 14

it gets there by "simplifying" the rational, so-called

Answer 15

\(\bf \large {
2\sqrt{2}sin(q)+2=0; 0 < q < 2\pi\\
sin(q) = -\frac{2}{2\sqrt{2}} =
-\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} =
-\frac{\sqrt{2}}{2}
}
\)

Answer 16

Alright, none of my choices are negative though.

Answer 17

well, check your Unit Circle for such values

Answer 18

between 0 and \(\bf \large \pi \) there'd be 2 angles with that sine value

Answer 19

I'm sorry I have got to sound dumb but I don't know what a unit circle is.

Answer 20

All of my options have two fractions. I am not sure how to paste the pictures to here.

Answer 21

http://i.stack.imgur.com/r8uHr.gif <---

Answer 22

many Unit Circles online, you'd need one :)

Answer 23

Thank you! Now you said between 0 and pi there would be angles with the sign value.

Answer 24

yes, and at those angles, is where the equation holds, that is \(\bf 2\sqrt{2}sin(q)+2=0\)

Answer 25

So my answer would be \[2\sqrt{2}\]

Answer 26

oops i meant to flip those

Answer 27

\[\sqrt{2}/2\]

Answer 28

well, "q" is an angle unit, so the values where \(\bf \large sin(q) = -\frac{\sqrt{2}}{2}\) is what "q" equals to, there are 2 angles, one in the 3rd quadrant and another in the 4th quadrant

Answer 29

ok so one of my answers should be
\[7\pi/4\]

Answer 30

and the other will be found at 135deg
\[\pi/4\]

Answer 31

!I see it. Thank you so much I actually understand how to do these now!!!