Question

solve number 23

Answer 1

@freckles

Answer 2

hey!

Answer 3

so we want to solve sin(x)=0.637
there are two quadrants in the first rotation that is the interval [0,2pi]
for which sin(x) is positive
that is in the 1st and 2nd quadrant
so you will have two answers to sin(x)=0.637

Answer 4

Hi

Answer 5

ok

Answer 6

so are they 3pi/2 and pi/2?

Answer 7

well 0.637 isn't a pretty number you can find on the unit circle
\[x=\arcsin(0.637) \text{ gives one solution }\]


now let's look at solving an example problem....
say we want to solve sin(x)=1/2
we can first solutions in first and second quadrant
we notice by looking at the unit circle we have x=pi/6 or x=5pi/6
let's look at at pi/6 and determine how we can get 5pi/6
well if you notice if you follow the line that is created by theta=5pi/6
we see we can also call that line theta=-pi/6
so if we change x=pi/6 to x=-pi/6+pi we will get out other solution
so lets work backwards and see what that show look like
so we know doing arcsin() of both sides of sin(x)=1/2
gives us the solution x=arcsin(1/2) or x=pi/6

so wherver I see pi/6 I'm going to write arcsin(1/2)

so we had the solution x=-pi/6+pi
so I'm going to write that as x=-arcsin(1/2)+pi
subtract pi on both sides
x-pi=-arcsin(1/2)
multiply both sides by -1
pi-x=arcsin(1/2)
so sin( ) of both sides gives
sin(pi-x)=1/2

Answer 8

So in your case you have the following two equations to solve:
\[\sin(x)=0.637 \text{ or } \sin(\pi-x)=0.637\]

Answer 9

so what I'm saying is you will get both of your solutions by solving or evaluating the following:
\[x=\arcsin(0.637) \text{ or } \pi-x=\arcsin(0.637)\]

Answer 10

ok

Answer 11

I got 2.451 for the second one

Answer 12

0.691 for the first one

Answer 13

so dats my answer?

Answer 14

that is what I have as well

Answer 15

k

Answer 16

well it was wrong

Answer 17

checking answers:
http://www.wolframalpha.com/input/?i=sin%282.451%29
http://www.wolframalpha.com/input/?i=sin%280.691%29
approximations look great

Answer 18

well I don't knw why it was marked wrong

Answer 19

anyways thank u for trying I would just ask my instructor tomorrow

Answer 20

just curious are u in college?

Answer 21

not anymoer

Answer 22

ok cool

Answer 23

at least u are free frm assignments, instructors e.t.c

Answer 24

I wonder if there is like a rounding thing

Answer 25

you know I know you round to 3 decimal places
but I wonder if the issue occurred in the rounding somewherw

Answer 26

mayb

Answer 27

hi i really hat to ask this but may you help me after?

Answer 28

but d rounding was right

Answer 29

like arcsin(0.637) gives 0.6906002396966 blah blah
rounded to 3 decimal places gives 0.691
as you said

I wonder if we also should round pi too 3 decimal places before using it
pi is approx 3.142

so we have
\[\sin(\pi-x)=0.637 \\ 3.142-x=\arcsin(0.637) \\ 3.142-x=0.691 \\ -x=0.691-3.142 \\ x=3.142-0.691=2.451\]
hmm so doing 3 digit rounding all the way throughout isn't the problem

Answer 30

yea

Answer 31

I wonder if the software finds the 0 before the decimal place significant
like instead of writing .691 write 0.691
so you would write 0.691 and 2.451 as the solutions

Answer 32

oh you already have in your solution

Answer 33

yes

Answer 34

yep I don't know but those are approximate the right solutions as we seen when we checked them
and I assume they wanted us to find the solutions in the interval [0,pi] since it said [0,pi]

Answer 35

yep

Answer 36

http://www.wolframalpha.com/input/?i=solve+sin%28x%29%3D0.637+on+%5B0%2C3.14%5D
look wolfram agrees with us

Answer 37

I will stop trying to argue for my answer now :p

Answer 38

well its ok

Answer 39

hey @diamondboy how many tries do you get?

Answer 40

if you get another try
this will probably not work but you can try
0.69 and 2.45

Answer 41

ok

Answer 42

I only have 5 tries and dat was d last one

Answer 43

ok well much sure you get your points back and argue with your teacher about it :p

Answer 44

make sure*

Answer 45

:) ok maybe not argue but wuld talk to her abt it

Answer 46

lol don't mean argue argue
politely defend your solution