At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

The intervals are:
\[0 \le x < 2\pi \]

and the second one is:
\[-\infty < x < \infty \]

Thanks for your reply! Can you explain why you add 2pi for the 1st one and 2npi for the second one?

Do you get it?

oh i see. so i have another question if you can help?

okay so 1.911-pi is the only answer

yeah sorry forgot to include that part but i got it

gosh you're a life saver!! could you check my work for this one as well?

sorry I made mistake

x=1.911

this one solution

cot x= -1 for -infinity< x < infinity
cosx/sinx= -1
3pi/4+ 2pi and 7pi/4+ 2pi ?

sorry i meant 3pi/4+ npi
7pi/4 + 4pi

7pi/4+ npi

for the previous problem the solution is x=1.911

yeah because you don't subtract possibilities right?

for the last on :x= 3pi/4+ 2npi and 7pi/4+ 2npi

*one

is it 2npi? i thought it was just npi because the period of cotangent is pi?

cot x=cos x/sin x, so it is contained on two functions, and period of cos x and sin x is 2npi

\[2\sin (2\theta)+ \sqrt{3}=0?\]

yeah sorry it got typed awkwardly

let solve it

I got (-pi/6)+npi, (-pi/3)+npi

this in general without considering the interval

could you explain? i think it might be because I didn't get rid of the 2 in sin2x= -sqrt(3)/2

your answer is also correct.

which one is correct?

is this right?
sin2x= \[-\sqrt{3}/2 \]
2x= 4pi/3 or 2x= 5pi/3
x= 2pi/3 or 5pi/6

correct and there are two more solutions.

2 more? I have no idea how to get them then

because npi

and the interval [0, 2pi)

I thought it was supposed to be 2npi? and i tried that and it went over 2pi

2x= 4pi/3 +2npi, 5pi/3+2npi
x= 2pi/3 +npi, 5pi/6+npi

Oh I didn't know you added the 2npi before simplifying

I forget it, you should and it

I need to eat my dinner. can we complete it later

sure! thank you so much for your help!!! i'll work on other things until then