Question

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!

Answer 1

do you have a unit circle with you?

Answer 2

\(\bf cos(x)=\cfrac{1}{2}\implies cos^{-1}[cos(x)]=cos^{-1}\left( \cfrac{1}{2} \right)\implies x=cos^{-1}\left( \cfrac{1}{2} \right)
\)

Answer 3

hmmm just notice is negative anyway \(\bf cos(x)=-\cfrac{1}{2}\implies cos^{-1}[cos(x)]=cos^{-1}\left( -\cfrac{1}{2} \right)\implies x=cos^{-1}\left( -\cfrac{1}{2} \right)
\)

Answer 4

The whole equation you replied with isn't showing up... @jdoe0001 But, I pulled up a unit circle.

Answer 5

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?

Answer 6

240 degrees and 120 degrees? With the radian values being 4pi/3 and 2pi/3 respectively? @jim_thompson5910

Answer 7

very good

Answer 8

those are 2 of infinitely many angle values x that make cos(x) = -1/2 true

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

yes it is

Answer 13

And could you help me with 1 more?

Answer 14

sure

Answer 15

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)

Answer 16

#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

Answer 17

#2 is incorrect

you use sin(A-B) = sin(A)cos(B) - cos(A)sin(B)

Answer 18

#3 is correct

cos(A-B) = cos(A)cos(B) + sin(A)sin(B)

Answer 19

So for#2, instead of sin9x, would it be sin7x?

Answer 20

yes

Answer 21

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 +?

Answer 22

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

Answer 23

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