A community for students.
Here's the question you clicked on:
 0 viewing
cassieforlife5
 one year ago
Solve the equation in the following intervals:
sin x = 0.2
I know for this you have to find sin^1 (0.2)
and I got 0.201 (my teacher said we have to do it in radians not degrees) but i'm so confused from here! Can anyone explain how to solve it?
cassieforlife5
 one year ago
Solve the equation in the following intervals: sin x = 0.2 I know for this you have to find sin^1 (0.2) and I got 0.201 (my teacher said we have to do it in radians not degrees) but i'm so confused from here! Can anyone explain how to solve it?

This Question is Closed

cassieforlife5
 one year ago
Best ResponseYou've already chosen the best response.0The intervals are: \[0 \le x < 2\pi \]

cassieforlife5
 one year ago
Best ResponseYou've already chosen the best response.0and the second one is: \[\infty < x < \infty \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[x=0.201358+2n \Pi\] for \[0 \le x < 2\pi\]: x=−0.201358+2Π for \[\infty < x < \infty\]: x=−0.201358+2nΠ, \[n=0,\pm1,\pm2,\pm3,....\]

cassieforlife5
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for your reply! Can you explain why you add 2pi for the 1st one and 2npi for the second one?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0because the period of sin x is 2Pi, it repeat it self after 2pi.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for 0≤x<2π : x=−0.201358+2Π this because the interval, −0.201358<0 so this solution is rejected x=−0.201358+2Π is good x=−0.201358+2*2Π>2Π this solution is rejected

cassieforlife5
 one year ago
Best ResponseYou've already chosen the best response.0oh i see. so i have another question if you can help?

cassieforlife5
 one year ago
Best ResponseYou've already chosen the best response.0sec x= 3 for the interval \[\pi \le x < \pi \] 1/cosx=3 cosx= 1/3 cos^1 (1/3)= 1.911 + pi is this the only answer or are there more?

cassieforlife5
 one year ago
Best ResponseYou've already chosen the best response.0okay so 1.911pi is the only answer

cassieforlife5
 one year ago
Best ResponseYou've already chosen the best response.0yeah sorry forgot to include that part but i got it

cassieforlife5
 one year ago
Best ResponseYou've already chosen the best response.0gosh you're a life saver!! could you check my work for this one as well?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sorry I made mistake

cassieforlife5
 one year ago
Best ResponseYou've already chosen the best response.0cot x= 1 for infinity< x < infinity cosx/sinx= 1 3pi/4+ 2pi and 7pi/4+ 2pi ?

cassieforlife5
 one year ago
Best ResponseYou've already chosen the best response.0sorry i meant 3pi/4+ npi 7pi/4 + 4pi

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for the previous problem the solution is x=1.911

cassieforlife5
 one year ago
Best ResponseYou've already chosen the best response.0yeah because you don't subtract possibilities right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for the last on :x= 3pi/4+ 2npi and 7pi/4+ 2npi

cassieforlife5
 one year ago
Best ResponseYou've already chosen the best response.0is it 2npi? i thought it was just npi because the period of cotangent is pi?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0cot x=cos x/sin x, so it is contained on two functions, and period of cos x and sin x is 2npi

cassieforlife5
 one year ago
Best ResponseYou've already chosen the best response.0oh... alright. sorry for asking so many questions but honestly you made it so much easier!! 2sin(2theta)+\[\sqrt{3}\]= 0 [0, 2pi) I got 4pi/3 and 5pi/3. It says to give both general and specific solutions, but these were the only ones that fit the intervals

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[2\sin (2\theta)+ \sqrt{3}=0?\]

cassieforlife5
 one year ago
Best ResponseYou've already chosen the best response.0yeah sorry it got typed awkwardly

cassieforlife5
 one year ago
Best ResponseYou've already chosen the best response.02sin2x=  sqrt(3) sin2x=  sqrt(3)/2 at 4pi/3 and 5pi/3 I've added 2npi, but all the solutions were larger than 2pi so I thought they were invalid

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got (pi/6)+npi, (pi/3)+npi

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this in general without considering the interval

cassieforlife5
 one year ago
Best ResponseYou've already chosen the best response.0could you explain? i think it might be because I didn't get rid of the 2 in sin2x= sqrt(3)/2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0your answer is also correct.

cassieforlife5
 one year ago
Best ResponseYou've already chosen the best response.0which one is correct?

cassieforlife5
 one year ago
Best ResponseYou've already chosen the best response.0is this right? sin2x= \[\sqrt{3}/2 \] 2x= 4pi/3 or 2x= 5pi/3 x= 2pi/3 or 5pi/6

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0correct and there are two more solutions.

cassieforlife5
 one year ago
Best ResponseYou've already chosen the best response.02 more? I have no idea how to get them then

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and the interval [0, 2pi)

cassieforlife5
 one year ago
Best ResponseYou've already chosen the best response.0I thought it was supposed to be 2npi? and i tried that and it went over 2pi

anonymous
 one year ago
Best ResponseYou've already chosen the best response.02x= 4pi/3 +2npi, 5pi/3+2npi x= 2pi/3 +npi, 5pi/6+npi

cassieforlife5
 one year ago
Best ResponseYou've already chosen the best response.0Oh I didn't know you added the 2npi before simplifying

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I forget it, you should and it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for the interval [0, 2pi): \[x=\frac{ 2\pi }{ 3 } , \frac{ 2\pi }{ 3 }+\pi=\frac{ 5\pi }{ 3 },\frac{ 5\pi }{ 6 } ,\frac{ 5\pi }{ 6 }+\pi=\frac{ 11\pi }{ 6}\]

cassieforlife5
 one year ago
Best ResponseYou've already chosen the best response.0yup I checked all 4 answers and they all equalled 0!! 2sin^2x= 2+cosx [0,pi] after using an identity and simplifying, I got cosx(2cosx +1)=0 cosx=0 at pi/2 and 3pi/2 2cosx +1> cosx= 1/2 at 2pi/3 and 4pi/3 I eliminated 3pi/2 and 4pi/3 because they were bigger than pi so my answers are pi/2 and 2pi/3

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I need to eat my dinner. can we complete it later

cassieforlife5
 one year ago
Best ResponseYou've already chosen the best response.0sure! thank you so much for your help!!! i'll work on other things until then
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.