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?

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.

A community for students.

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?

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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 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: \[-\infty < x < \infty \]
\[x=-0.201358+2n \Pi\] for \[0 \le x < 2\pi\]: x=−0.201358+2Π for \[-\infty < x < \infty\]: x=−0.201358+2nΠ, \[n=0,\pm1,\pm2,\pm3,....\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Thanks for your reply! Can you explain why you add 2pi for the 1st one and 2npi for the second one?
because the period of sin x is 2Pi, it repeat it self after 2pi.
1 Attachment
for 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
Do you get it?
oh i see. so i have another question if you can help?
sec 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?
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
oh... 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
\[2\sin (2\theta)+ \sqrt{3}=0?\]
yeah sorry it got typed awkwardly
let solve it
2sin2x= - 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
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
for the interval [0, 2pi): \[x=\frac{ 2\pi }{ 3 } , \frac{ 2\pi }{ 3 }+\pi=\frac{ 5\pi }{ 3 },\frac{ 5\pi }{ 6 } ,\frac{ 5\pi }{ 6 }+\pi=\frac{ 11\pi }{ 6}\]
yup 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
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

Not the answer you are looking for?

Search for more explanations.

Ask your own question