Question

Solve each equation in the interval [0,2pi) rounded to two decimal points
a) 3sin(theta)-1=0

Answer 1

I got
sin(theta)=1/3
theta = 19.47
but this is not correct, am I meant to be even solving for theta?

Answer 2

maybe you are supposed to be answering in radians, not degrees
who knows?

Answer 3

hrrm yeah doing it in rad gives me the correct answer =.= but they never specified that. sigh lol

Answer 4

http://www.wolframalpha.com/input/?i=arcsin%281%2F3%29

Answer 5

yeah the answers they give are 0.34 and 2.80
where is the 2.80 from?

Answer 6

for a second pretend we want to solve this one on [0,2pi)

\[\sin(\theta)=\frac{1}{2} \]
The unit circle gives the solutions as:
\[\theta=\frac{\pi}{6}, \frac{5\pi}{6}\]
The pi/6 can be obtain by taking arcsin( ) of both sides

\[\theta=\arcsin(\frac{1}{2}) =\frac{\pi}{6} \approx 0.52 \\ \text{ now how we can get the other solution without looking at the unit circle } \\ \\ \text{ well remember } \sin \text{ is odd } \sin(\theta)=-\sin(-\theta) \\ \sin(\frac{\pi}{6})=-\sin(-\frac{\pi}{6}) \\ \text{ also recall \sum identity for } \sin \\ \sin(\pi-\frac{\pi}{6}) =\sin(\pi)\cos(\frac{\pi}{6})-\sin(\frac{\pi}{6})\cos(\pi) \\ =\sin(\pi-\frac{\pi}{6})=0 \cdot \cos(\frac{\pi}{6})-(-1)\sin(\frac{\pi}{6}) \\ =\sin(\pi-\frac{\pi}{6})=-(-1)\sin(\frac{\pi}{6}) \\ \text{ now recall } -\sin(\frac{\pi}{6})=\sin(-\frac{\pi}{6}) \\ \text{ so we have } \sin(\frac{\pi}{6})=\sin(\pi-\frac{\pi}{6}) =\sin(\frac{6\pi}{6}-\frac{\pi}{6})=\sin(\frac{5\pi}{6})\]
so the other solution we can said is given by the equation:
\[-\theta+\pi= \arcsin(\frac{1}{2}) \\ \theta-\pi=-\arcsin(\frac{1}{2}) \\ \theta=\pi-\arcsin(\frac{1}{2})=\pi-\frac{\pi}{6}=\frac{5\pi}{6}\]

so what I am saying for you:
you have the following two solutions:
\[\theta=\arcsin(\frac{1}{3}) \text{ or } -\theta+\pi=\arcsin(\frac{1}{3}) \\ \theta=\arcsin(\frac{1}{3}) \text{ or } \theta=\pi-\arcsin(\frac{1}{3}) \\ \theta \ \]

Answer 7

To help guide me I kinda look at the unit circle sometimes to remember me of the equations I need to solve things like this

Answer 8

oh gosh this is going to take a while to get through to me haha but yeah it makes sense :)

Answer 9

adding two pi to
theta = ___ -arcsin(1/3)
would be out of restrictions, correct? so that's why we only add pi?

Answer 10

looking back at theta=___ -arcsin(1/2)
we know arcsin(1/2) is pi/6
so we are looking at theta=___-pi/6
now putting 2pi in that blank
would actually take us back to -pi/6 instead of pi/6

Answer 11

look at the unit circle for a second
let's look at the pi/6 and -pi/6
sin(pi/6)=1/2
where as
sin(-pi/6)=-1/2
but this isn't totally bad because if you stay on that line at -pi/6
....and move your finger on that line where -pi/6 is ...you will see the desire solution of sin being 1/2 again which would be at 5pi/6

and guess what the angle created by a line is 180 deg

or you would say pi
and pi-pi/6 would be another name for that 5pi/6

Answer 12

so to solve:
\[\sin(\theta)=a \text{ for } \theta \text{ pretending } a \text{ is within range of } \sin( ) \\ \theta=\arcsin(a) \text{ will give one solution in } [0,2\pi] \\ \theta=\pi-\arcsin(a) \text{ will give another solution in } [0,2\pi] \\ \text{ this will always be two solutions if } \arcsin(a) \text{ isn't} \frac{\pi}{2} \text{ or } \frac{-\pi}{2} \\ \text{ because in those cases there will only be one solution in } [0,2\pi]\]

Answer 13

hmm hrm. Ok. I'll have to do a few more and play around with it, but i'm kindd of getting the hang of it o.o

Answer 14

so you understand that sin(u)=2 does not satisfy what I said above right?

Answer 15

do you know why?

Answer 16

It's out of domain? o_o

Answer 17

hells yes
sin ranges from -1 to 1
this would I mean pretending a is within in the range of sin( )
very good

Answer 18

\[\cos(\theta)=a \text{ pretending } a \text{ is within range of } \cos( ) \\ \theta=\arccos(a) \text{ will give one solution in } [0,2\pi] \\ \theta=-\arccos(a)+2\pi \text{ will give another solution in } [0,2\pi ] \\ \text{ this will always be two solutions if } \arccos(a) \text{ isn't } \pi \\ \text{ because this will give only one solution}\]

Answer 19

All of the generalizations I made are only for [0,2pi]

Answer 20

....haha ai .

Answer 21

oh. right right.

Answer 22

Phew haha
dude, you're amazing. You've stuck with me and my ignorance for so long xD

Answer 23

it is kinda of fun :)
we could easily extend what I have said so you all the real solutions in the whole world

Answer 24

\[\cos(\theta)=a \text{ where } -1 \le a \le 1 \text{ has solutions } \\ \theta=\arccos(a)+2 \pi n \\ \theta=-\arccos(a)+2 \pi n \\ \text{ where } n \text{ is integer } \\ \sin(\theta)=a \text{ where} -1 \le a \le 1 \text{ has solutions } \\ \theta=\arcsin(a)+2 \pi n \\ \theta=\pi-\arcsin(a)+2 \pi n =-\arcsin(a)+\pi(1+2n) \\ \text{ where} n \text{ is integer }\]

Answer 25

you just add 2 pi n to the solutions since both sin and cos have period 2pi :)

Answer 26

it would go on forever, infinite answers haha