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

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Babynini
 one year ago
Best ResponseYou've already chosen the best response.1I got sin(theta)=1/3 theta = 19.47 but this is not correct, am I meant to be even solving for theta?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0maybe you are supposed to be answering in radians, not degrees who knows?

Babynini
 one year ago
Best ResponseYou've already chosen the best response.1hrrm yeah doing it in rad gives me the correct answer =.= but they never specified that. sigh lol

Babynini
 one year ago
Best ResponseYou've already chosen the best response.1yeah the answers they give are 0.34 and 2.80 where is the 2.80 from?

freckles
 one year ago
Best ResponseYou've already chosen the best response.0for 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: \[\theta=\arcsin(\frac{1}{3}) \text{ or } \theta+\pi=\arcsin(\frac{1}{3}) \\ \theta=\arcsin(\frac{1}{3}) \text{ or } \theta=\pi\arcsin(\frac{1}{3}) \\ \theta \ \]

freckles
 one year ago
Best ResponseYou've already chosen the best response.0To help guide me I kinda look at the unit circle sometimes to remember me of the equations I need to solve things like this

Babynini
 one year ago
Best ResponseYou've already chosen the best response.1oh gosh this is going to take a while to get through to me haha but yeah it makes sense :)

Babynini
 one year ago
Best ResponseYou've already chosen the best response.1adding two pi to theta = ___ arcsin(1/3) would be out of restrictions, correct? so that's why we only add pi?

freckles
 one year ago
Best ResponseYou've already chosen the best response.0looking 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.0look 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 pipi/6 would be another name for that 5pi/6

freckles
 one year ago
Best ResponseYou've already chosen the best response.0so to solve: \[\sin(\theta)=a \text{ for } \theta \text{ pretending } a \text{ is within range of } \sin( ) \\ \theta=\arcsin(a) \text{ will give one solution in } [0,2\pi] \\ \theta=\pi\arcsin(a) \text{ will give another solution in } [0,2\pi] \\ \text{ this will always be two solutions if } \arcsin(a) \text{ isn't} \frac{\pi}{2} \text{ or } \frac{\pi}{2} \\ \text{ because in those cases there will only be one solution in } [0,2\pi]\]

Babynini
 one year ago
Best ResponseYou've already chosen the best response.1hmm 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.0so you understand that sin(u)=2 does not satisfy what I said above right?

Babynini
 one year ago
Best ResponseYou've already chosen the best response.1It's out of domain? o_o

freckles
 one year ago
Best ResponseYou've already chosen the best response.0hells yes sin ranges from 1 to 1 this would I mean pretending a is within in the range of sin( ) very good

freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[\cos(\theta)=a \text{ pretending } a \text{ is within range of } \cos( ) \\ \theta=\arccos(a) \text{ will give one solution in } [0,2\pi] \\ \theta=\arccos(a)+2\pi \text{ will give another solution in } [0,2\pi ] \\ \text{ this will always be two solutions if } \arccos(a) \text{ isn't } \pi \\ \text{ because this will give only one solution}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.0All of the generalizations I made are only for [0,2pi]

Babynini
 one year ago
Best ResponseYou've already chosen the best response.1Phew haha dude, you're amazing. You've stuck with me and my ignorance for so long xD

freckles
 one year ago
Best ResponseYou've already chosen the best response.0it is kinda of fun :) we could easily extend what I have said so you all the real solutions in the whole world

freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[\cos(\theta)=a \text{ where } 1 \le a \le 1 \text{ has solutions } \\ \theta=\arccos(a)+2 \pi n \\ \theta=\arccos(a)+2 \pi n \\ \text{ where } n \text{ is integer } \\ \sin(\theta)=a \text{ where} 1 \le a \le 1 \text{ has solutions } \\ \theta=\arcsin(a)+2 \pi n \\ \theta=\pi\arcsin(a)+2 \pi n =\arcsin(a)+\pi(1+2n) \\ \text{ where} n \text{ is integer }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.0you just add 2 pi n to the solutions since both sin and cos have period 2pi :)

Babynini
 one year ago
Best ResponseYou've already chosen the best response.1it would go on forever, infinite answers haha
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