cassieforlife5
  • cassieforlife5
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
cassieforlife5
  • cassieforlife5
The intervals are: \[0 \le x < 2\pi \]
cassieforlife5
  • cassieforlife5
and the second one is: \[-\infty < x < \infty \]
anonymous
  • anonymous
\[x=-0.201358+2n \Pi\] for \[0 \le x < 2\pi\]: x=−0.201358+2Π for \[-\infty < x < \infty\]: x=−0.201358+2nΠ, \[n=0,\pm1,\pm2,\pm3,....\]

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cassieforlife5
  • cassieforlife5
Thanks for your reply! Can you explain why you add 2pi for the 1st one and 2npi for the second one?
anonymous
  • anonymous
because the period of sin x is 2Pi, it repeat it self after 2pi.
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anonymous
  • anonymous
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
anonymous
  • anonymous
Do you get it?
cassieforlife5
  • cassieforlife5
oh i see. so i have another question if you can help?
cassieforlife5
  • cassieforlife5
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?
cassieforlife5
  • cassieforlife5
okay so 1.911-pi is the only answer
cassieforlife5
  • cassieforlife5
yeah sorry forgot to include that part but i got it
cassieforlife5
  • cassieforlife5
gosh you're a life saver!! could you check my work for this one as well?
anonymous
  • anonymous
sorry I made mistake
anonymous
  • anonymous
x=1.911
anonymous
  • anonymous
this one solution
cassieforlife5
  • cassieforlife5
cot x= -1 for -infinity< x < infinity cosx/sinx= -1 3pi/4+ 2pi and 7pi/4+ 2pi ?
cassieforlife5
  • cassieforlife5
sorry i meant 3pi/4+ npi 7pi/4 + 4pi
cassieforlife5
  • cassieforlife5
7pi/4+ npi
anonymous
  • anonymous
for the previous problem the solution is x=1.911
cassieforlife5
  • cassieforlife5
yeah because you don't subtract possibilities right?
anonymous
  • anonymous
for the last on :x= 3pi/4+ 2npi and 7pi/4+ 2npi
anonymous
  • anonymous
*one
cassieforlife5
  • cassieforlife5
is it 2npi? i thought it was just npi because the period of cotangent is pi?
anonymous
  • anonymous
cot x=cos x/sin x, so it is contained on two functions, and period of cos x and sin x is 2npi
cassieforlife5
  • cassieforlife5
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
anonymous
  • anonymous
\[2\sin (2\theta)+ \sqrt{3}=0?\]
cassieforlife5
  • cassieforlife5
yeah sorry it got typed awkwardly
anonymous
  • anonymous
let solve it
cassieforlife5
  • cassieforlife5
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
anonymous
  • anonymous
I got (-pi/6)+npi, (-pi/3)+npi
anonymous
  • anonymous
this in general without considering the interval
cassieforlife5
  • cassieforlife5
could you explain? i think it might be because I didn't get rid of the 2 in sin2x= -sqrt(3)/2
anonymous
  • anonymous
your answer is also correct.
cassieforlife5
  • cassieforlife5
which one is correct?
cassieforlife5
  • cassieforlife5
is this right? sin2x= \[-\sqrt{3}/2 \] 2x= 4pi/3 or 2x= 5pi/3 x= 2pi/3 or 5pi/6
anonymous
  • anonymous
correct and there are two more solutions.
cassieforlife5
  • cassieforlife5
2 more? I have no idea how to get them then
anonymous
  • anonymous
because npi
anonymous
  • anonymous
and the interval [0, 2pi)
cassieforlife5
  • cassieforlife5
I thought it was supposed to be 2npi? and i tried that and it went over 2pi
anonymous
  • anonymous
2x= 4pi/3 +2npi, 5pi/3+2npi x= 2pi/3 +npi, 5pi/6+npi
cassieforlife5
  • cassieforlife5
Oh I didn't know you added the 2npi before simplifying
anonymous
  • anonymous
I forget it, you should and it
anonymous
  • anonymous
for the interval [0, 2pi): \[x=\frac{ 2\pi }{ 3 } , \frac{ 2\pi }{ 3 }+\pi=\frac{ 5\pi }{ 3 },\frac{ 5\pi }{ 6 } ,\frac{ 5\pi }{ 6 }+\pi=\frac{ 11\pi }{ 6}\]
cassieforlife5
  • cassieforlife5
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
anonymous
  • anonymous
I need to eat my dinner. can we complete it later
cassieforlife5
  • cassieforlife5
sure! thank you so much for your help!!! i'll work on other things until then

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