Question

once again, i need help please!

(cos x)(cot x) + (cos x) = 0

solve over 0, 2pi

Answer 1

okay, so far i got this:

(cos x)(cot x) + cos x = 0
cos x * cot x = - cos x
cot x = -cos x/cos x
cot x = -1

am I doing this right so far? XD

Answer 2

Lookin' good!

Answer 3

so, if i did this right, it means that 135 degrees and 315 degrees angles, or 3pi/4 and 7pi/4, is the answer? because tanget is -1, so cot would flip -1/1 so it'd still be -1, right?

Answer 4

(cos x)(cot x) + cos x = 0

cos x (cot x +1) = 0

cos x = 0 and cot x+1 =0
you need to consider both of these equations :)

Answer 5

because your original equation will get satisfied even when cos x =0, right ?

Answer 6

cos x = 0, it doesn't work. becuase that means itll be 90 and 270 respectively

tan of 90 is undefined, and cot is 0
tan of 270 is undefined and cot is 0 still.

thnaks ofr hte help though, it made me second guess myself

Answer 7

cos 90 times cot 90 is still 0

so they both do work :)

Answer 8

my instructor told me i have to idenify the variables by themselves, so i have to evaluate cot x = -1 and cos x = 0, individually.

Answer 9

\[ cosx (cosx/sinx)+cosx=0 <=> cos^{2}x+cosxsinx=0 <=> cosx(cosx+sinx)=0 \]
(i) \[ <=> cosx=0 <=> cosx=cosπ/2 <=> x=2kπ\pmπ/2 \]
but \[0\le x\le 2π <=>0\le 2kπ\pmπ/2\le2π <=> 0\le k \pm1/4 \le1\]
\[ <=> \mp1/4\le k\le1\mp1/4\]
but \[k \in \mathbb{Z}\]
so k=0 or k=1
\[x=\pmπ/2 \] or \[x=5π/2\] or \[x=3π/2\]
and (ii)\[ cosx +sinx=0\]
\[ cosx=-sinx<=> 1=-tanx (*) <=>tanx=-1 <=> tanx=tan7π/4 \]
\[<=>x=kπ+7π/4\]
but \[x \in [0,2π] \]
\[0\le kπ+7π/4\le 2π <=> -7/4\le k\le1/4\]
so k=-1,0
k=-1, x=3π/4
k=0, x=7π/4
(*) \[cosx \neq 0 \] so (i) step is cancelled
So x=3π/4 and x=7π/4