how to solve cos (4x)=-1/2 ?
find a number (angle if you like) whose cosine is \(-\frac{1}{2}\) which should not be too hard
if it is not obvious, look at the unit circle on the last page of the attached cheat sheet see what angle has a first coordinate of \(-\frac{1}{2}\)
wouldnt that be both pi/3 and 5pi/3??? would this yield two general answers?
no \(\cos(\frac{\pi}{3})=\frac{1}{2}\) you want \(-\frac{1}{2}\)
try \(\frac{2\pi}{3}\) and \(\frac{4\pi}{3}\)
this tells you \[4x=\frac{2\pi}{3}\] and so \[x=\frac{\pi}{6}\]
or \[4x=\frac{4\pi}{3}\] and therefore \[x=\frac{\pi}{3}\]
oh ya! so then it still gives the answers of 2pi/3 +2pi n and 4pi/3 +2pi n ?
wat about the other value of x?
no you are solving for \(x\) not for \(4x\)
gotta divide your answers by 4
\[4x=\frac{2\pi}{3}+2n\pi\] \[x=\frac{\pi}{6}+\frac{1}{2}n\pi\]
I mean like cos(2pi/3) and cos(4pi/3)...
would u also plug in the 4pi/3, i mean.
hold on lets go slow
\[\cos (\frac{2\pi}{3})=-\frac{1}{2}\] right?
yes
and your job was to solve \[\cos(4x)=-\frac{1}{2}\]
yes
this tells you that \[4x=\frac{2\pi}{3}+2n\pi\]
yea
now divide by \(4\) to solve for \(x\) and get \[x=\frac{\pi}{6}+\frac{1}{2}n\pi\]
or however you wish to write it
So my question is,
are there two answers, since there are two different angles that cosine can be if it's -1/2?
right now repeat the process with \[4x=\frac{4\pi}{3}+2n\pi\]
yes there are two answers, or rather two formulas for infinite number of answers
pi/6 + pi n/2 and pi/3 + pi n/2 yes? :)
yes!
>o< ok then that clears things up~ and the +2pi n thing would be just +pi n for tangent and cotangent, right? I just need to clear that up...
yes
is there a plus or minus thing involved in solving trig equations? I went in to tutoring this morning and heard something about having to put plus or minus in the process, but I had to go early, so I wasn't able to ask what scenario that was in... I think I heard something about cosine...
hello?
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