anonymous
  • anonymous
Could somebody help me with the question: Find all the solutions for the equation: cosx = -1/2 Not looking for straight answers, but verification of the answer after an explanation on how to solve? Thank you!
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
jim_thompson5910
  • jim_thompson5910
do you have a unit circle with you?
jdoe0001
  • jdoe0001
\(\bf cos(x)=\cfrac{1}{2}\implies cos^{-1}[cos(x)]=cos^{-1}\left( \cfrac{1}{2} \right)\implies x=cos^{-1}\left( \cfrac{1}{2} \right) \)
jdoe0001
  • jdoe0001
hmmm just notice is negative anyway \(\bf cos(x)=-\cfrac{1}{2}\implies cos^{-1}[cos(x)]=cos^{-1}\left( -\cfrac{1}{2} \right)\implies x=cos^{-1}\left( -\cfrac{1}{2} \right) \)

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
The whole equation you replied with isn't showing up... @jdoe0001 But, I pulled up a unit circle.
jim_thompson5910
  • jim_thompson5910
look at the points on the unit circle that have an x coordinate of -1/2 what are the angles that correspond to these points?
anonymous
  • anonymous
240 degrees and 120 degrees? With the radian values being 4pi/3 and 2pi/3 respectively? @jim_thompson5910
jim_thompson5910
  • jim_thompson5910
very good
jim_thompson5910
  • jim_thompson5910
those are 2 of infinitely many angle values x that make cos(x) = -1/2 true
jim_thompson5910
  • jim_thompson5910
add on 360 (in degree mode) or 2pi (radian mode) to get coterminal angles. You can also subtract 360 or 2pi to get other coterminal angles. There is no limit to how much you can add or subtract
jim_thompson5910
  • jim_thompson5910
So if you're in degree mode, then the solution set is x = 120 + 360*n or x = 240 + 360*n where n is an integer
anonymous
  • anonymous
So with the answer choices of: A. {2pi/3+npi | n = o. +/-1, +/-2, ... } B. {5pi/6+npi | n = o. +/-1, +/-2, ... } C. {5pi/6+2npi, 7pi/6+2npi | n = o. +/-1, +/-2, ... } D. {2pi/3+2npi , 4pi/3+2npi | n = o. +/-1, +/-2, ... } Would the answer be D? @jim_thompson5910
jim_thompson5910
  • jim_thompson5910
yes it is
anonymous
  • anonymous
And could you help me with 1 more?
jim_thompson5910
  • jim_thompson5910
sure
anonymous
  • anonymous
Well, three actually. I think I got them correct, but just verifying. 1. For cos94cos37 + sin94sin37 (degrees) would the answer be cos57? 2. For sin8xcosx - cos8xsinx, I got (in terms of sin) sin9x 3. And for cos8xcos2x - sin8xsin2x, I got cos10x Just confused, on those, when to use (cos a + b), (cos a - b), (sin a + b), or (sin a - b)
jim_thompson5910
  • jim_thompson5910
#1 is correct you can use a calculator to get cos(94)*cos(37)+sin(94)*sin(37) = 0.54463903501502 cos(57) = 0.54463903501502 both are equal to the same decimal value. You can also subtract the two values and you'll get 0 or very close to it
jim_thompson5910
  • jim_thompson5910
#2 is incorrect you use sin(A-B) = sin(A)cos(B) - cos(A)sin(B)
jim_thompson5910
  • jim_thompson5910
#3 is correct cos(A-B) = cos(A)cos(B) + sin(A)sin(B)
anonymous
  • anonymous
So for#2, instead of sin9x, would it be sin7x?
jim_thompson5910
  • jim_thompson5910
yes
anonymous
  • anonymous
Can you explain when to use which identity? Like when to use the sin difference or sum, vs. the cos difference or sum? And when to use - vs +?
jim_thompson5910
  • jim_thompson5910
when you have the two 'cos' terms together, like with cos(A)cos(B) + sin(A)sin(B), you use the cos(A+B) or cos(A-B) identity
jim_thompson5910
  • jim_thompson5910
whatever symbol is between the cos(A)cos(B) and sin(A)sin(B) is going to be the opposite inside the cosine on the left side
jim_thompson5910
  • jim_thompson5910
|dw:1436920932555:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1436920967189:dw|
jim_thompson5910
  • jim_thompson5910
and if you have sin mixed with cosine, like sin(2x)cos(x) + cos(x)sin(2x), then you'll use sin(A+B) or sin(A-B)
jim_thompson5910
  • jim_thompson5910
|dw:1436921078884:dw|
jim_thompson5910
  • jim_thompson5910
It comes down to recognizing the form. That happens with lots of practice and memorization. You can look up the identities on a reference sheet like this one http://tutorial.math.lamar.edu/pdf/Trig_Cheat_Sheet.pdf (see the "Sum and Difference Formulas" section on page 2) but sometimes the teacher won't let you have a reference sheet like that. It all depends.

Looking for something else?

Not the answer you are looking for? Search for more explanations.