Question

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?

Answer 1

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

Answer 2

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

Answer 3

\[x=-0.201358+2n \Pi\]
for \[0 \le x < 2\pi\]:
x=−0.201358+2Π
for
\[-\infty < x < \infty\]:
x=−0.201358+2nΠ, \[n=0,\pm1,\pm2,\pm3,....\]

Answer 4

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

Answer 5

because the period of sin x is 2Pi, it repeat it self after 2pi.

Answer 6

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

Answer 7

Do you get it?

Answer 8

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

Answer 9

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?

Answer 10

okay so 1.911-pi is the only answer

Answer 11

yeah sorry forgot to include that part but i got it

Answer 12

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

Answer 13

sorry I made mistake

Answer 14

x=1.911

Answer 15

this one solution

Answer 16

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

Answer 17

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

Answer 18

7pi/4+ npi

Answer 19

for the previous problem the solution is x=1.911

Answer 20

yeah because you don't subtract possibilities right?

Answer 21

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

Answer 22

*one

Answer 23

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

Answer 24

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

Answer 25

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

Answer 26

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

Answer 27

yeah sorry it got typed awkwardly

Answer 28

let solve it

Answer 29

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

Answer 30

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

Answer 31

this in general without considering the interval

Answer 32

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

Answer 33

your answer is also correct.

Answer 34

which one is correct?

Answer 35

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

Answer 36

correct and there are two more solutions.

Answer 37

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

Answer 38

because npi

Answer 39

and the interval [0, 2pi)

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

I forget it, you should and it

Answer 44

for the interval [0, 2pi):
\[x=\frac{ 2\pi }{ 3 } , \frac{ 2\pi }{ 3 }+\pi=\frac{ 5\pi }{ 3 },\frac{ 5\pi }{ 6 } ,\frac{ 5\pi }{ 6 }+\pi=\frac{ 11\pi }{ 6}\]

Answer 45

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

Answer 46

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

Answer 47

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