Let's start off easy :) trig help???
pi/18
i messed up my equation let me reqrite
\[\Huge \cos3x= -\frac{ 1 }{ 2 }\]
sorry
2pi/9
\Huge Also Solve , [0,2pi)
can you please explain why its that?
unit circle memorization. it shows us that cos-1(-1/2) is 2pi/3 and were looking for 1/3 of that since its 3x so therefore its 2pi/9 heres a unit circle http://www.google.com/imgres?imgurl=http://upload.wikimedia.org/wikipedia/commons/thumb/4/4c/Unit_circle_angles_color.svg/720px-Unit_circle_angles_color.svg.png&imgrefurl=http://en.wikipedia.org/wiki/Unit_circle&h=225&w=225&tbnid=NHFRVspjd-vZrM:&zoom=1&tbnh=186&tbnw=186&usg=__sOrTVeIoKl48_TodEan8dIx7UGk=&docid=D09vYD7H2hKF-M&itg=1&sa=X&ei=yjZLU-WNBMie2wXK2IDQDg&ved=0CIYBEPwdMAo
2pi/9 = -(1/2)???
(3)*2pi/9 =-1/2 you were looking for x and you had cos(3x) so yeah
o wow i understand -.-
u meant i shoud move cos to the other side.
boom... cos-(1/2) = 3x
\[\huge\frac{ 2\pi }{ 3 } * \frac{ 1 }{ 3 } = x\]
is that what you meant?
no because you have to take the inverse of cosine because cosine of 1/2 is not the same as cos(3x) memorize the unit circle because it makes calc a lot easier. you need to look for where cosine would be -1/2 which is at 2pi/3 and to solve for x from there you just need to divide 2pi/3 by 3
I'm sorry if this isn't making perfect sense but its been 4 years since I took precalc and I'm not sure how to word it other than "well that's the way it is"
i got x = 2pi/9 though lol
hmm
let me c
that is right though
peachy we did the same thing....................................................
-____________________-
we just worded it differently
but cosine of -1/2 is .87758 so when you wrote what you did I didn't follow your steps
"boom... cos-(1/2) = 3x" cuz this doesn't make sense
I don't understand now why did you go into decimals? o-o
ahhhh i shifted the cos from cos3x to the other side.^^^
yeah math won't let you do that directly
so it became 3x = cos -(1/2) from cos3x -(1/2)
oh
thats where we differ because the only way to do that is to take the inverse of cosine do you know what that is?
no please explain :D
cos(0)=1 so the inverse of cosine (or cos^-1(x) cos^-1(1) = 0
its just a way of flipping them
oh i get it
now how did u apply it to this equation again?
so to solve this youre going to wanna take the inverse cosine of -1/2
mhm
and when you look at the unit circle which i linked above you can see that cosine = -1/2 when cos(2pi/3) do you follow me so far
yes
inverse cosine of -1/2 is 120 = 2pi/3
you need to solve so 3x = 2pi/3 now
oh ok i seee ittt
to fit the equation you were given and then 2pi/9
divide by 3 3on both sides then we're golden
so you followed all of that?
yep thank you
feel free to message me for more help! sorry I took so long explaining =/
its fine you helped a lot.
\[\Huge \sin2x+sinx = 0\]
how would i go about solving this within [0,2pi)
sin2x = 2sinxcosx
yep so set that equal to -sin(x)
2sincosx = -sinx???
yes maam
divide a sin over
awesome thanks
then easy peasy from there right?
when i do identities and solving I always feel like i dont know it and am stuck but then someone tells me and im like OHHHHHHHHHHH i just dont remember the identity unless someone helps me recall it
snxcosx=1
make flashcards for identities it helps
wait am i right?
do i like square both sides or something
wait for the problem we were working on
yep we're now at sinxcosx = 1
no we had 2*sin(x)*cos(x)=-sin(x)
right?
yes
i wrote something wrong mb
i forgive you ;)
divide a sin over and we get 2cos(x)=-1
then cos(x)=-1/2
cosx = -1/2
girl why do you still have a sin there |dw:1397440087760:dw|
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