Question

\[\sin (2x) = -\sqrt{(2)}/(2) \]

Answer 1

Solve each equation for 0 <= x <2pi

Answer 2

\[\sin 2x=-\frac{ \sqrt{2} }{ 2 }=-\frac{ 1 }{ \sqrt{2} }=-\sin \frac{ \pi }{ 6 }=\sin \left( \pi+\frac{ \pi }{ 6 } \right),\sin (2 \pi- \frac{ \pi }{6 } ),\]
\[\sin \left( 2 \pi+\pi+\frac{ \pi }{ 6 } \right),\sin \left( 2 \pi+2 \pi-\frac{ \pi }{ 6 } \right)\]
2x=?
x=?

Answer 3

I don't really understand this at all. I do under the first 3 steps though.

Answer 4

understand*

Answer 5

Wifi dropped, sorry..

Answer 6

Is there only two solutions?

Answer 7

I know you can make a graph for these, but I have never made a sin(2x) graph

Answer 8

sorry i wrongly typed

Answer 9

\[2x=\frac{ 5 \pi }{ 4 },\frac{ 7 \pi }{ 4 },\frac{ 13 \pi }{ 4 },\frac{ 15 \pi }{ 4 }\]
x=?

Answer 10

Am I supposed to divide by 2?

Answer 11

write \[\frac{ \pi }{ 4 }~\in~place~of~\frac{ \pi }{ 6 } \]

Answer 12

yes divide by 2

Answer 13

I don't think I saw one before where it went from 5 (+2), 7 (+6), 13(+2), Isn't it a constant?

Answer 14

if it is 2x ,then add 2n pi,where n=0,1
if it is 3x,then add 2 n pi,where n=0,1,2
.....

Answer 15

I'm new to this subject, so I haven't fully grasped it yet.

Answer 16

\[\sin 2x=\sin \left( 2 n \pi+\frac{ 5 \pi }{ 4 } \right),\sin \left( 2n \pi+\frac{ 7 \pi }{ 4 } \right)\]

Answer 17

one answer is pi/4?

Answer 18

as sin 2x is negative so it lies in 3 rd and 4th quadrant.

Answer 19

All the solutions are in the 3rd and 4th?

Answer 20

\[\frac{ 5 \pi }{ 8 },\frac{ 7 \pi }{ 8 },\frac{ 13 \pi }{ 8 },\frac{ 15 \pi }{ 8 }\]

Answer 21

Does that mean there is 8 solutions?

Answer 22

no ,we have to take between 0 and 2 pi .
they are only four.

Answer 23

Oh, I see now.

Answer 24

Do you know what the graph looks like for this kind of equation?

Answer 25

remember 2x lies in 3rd and 4th quadrant.

Answer 26

So the negative part of the quadrant, yes?

Answer 27

sorry i have to go to bed now and leaving.