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

1. Babynini

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

2. anonymous

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

3. Babynini

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

4. anonymous
5. Babynini

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

6. freckles

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 \$

7. freckles

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

8. Babynini

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

9. Babynini

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

10. freckles

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

11. freckles

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

12. freckles

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

13. Babynini

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

14. freckles

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

15. freckles

do you know why?

16. Babynini

It's out of domain? o_o

17. freckles

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

18. freckles

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

19. freckles

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

20. Babynini

....haha ai .

21. Babynini

oh. right right.

22. Babynini

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

23. freckles

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

24. freckles

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

25. freckles

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

26. Babynini

it would go on forever, infinite answers haha

Find more explanations on OpenStudy