sammietaygreen
  • sammietaygreen
Find the values of x for which the equation cos x = -1 is true. k represents an integer.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
there is no k in your question
sammietaygreen
  • sammietaygreen
anonymous
  • anonymous
7

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More answers

sammietaygreen
  • sammietaygreen
I got the last one
SolomonZelman
  • SolomonZelman
cos(x)=-1 cos(x)+1=0 That means that your x-solutions are the roots of y=cos(x)+1 https://www.desmos.com/calculator/ahvced9xlt
sammietaygreen
  • sammietaygreen
What do you mean by roots
SolomonZelman
  • SolomonZelman
roots of the functions are the zeros (or the x-intercepts) of the function.
sammietaygreen
  • sammietaygreen
I still don't get it honestly. Do I have to find the zeros of the equation
sammietaygreen
  • sammietaygreen
@jim_thompson5910
jim_thompson5910
  • jim_thompson5910
for each answer choice, let's replace k with some whole number k = 0 is probably the easiest to deal with
jim_thompson5910
  • jim_thompson5910
so instead of 2pi*k, it is 2pi*k = 2pi*0 = 0 what happens when you plug x = 0 into the original equation?
sammietaygreen
  • sammietaygreen
well, cos(0) = 1
sammietaygreen
  • sammietaygreen
is this basically plugging in?
jim_thompson5910
  • jim_thompson5910
so `cos(x) = -1` turns into `1 = -1` when you replaced x with 0 and evaluated the left hand side
jim_thompson5910
  • jim_thompson5910
`1 = -1` is a false equation, so x = 0 is NOT a solution (it is not in the solution set)
sammietaygreen
  • sammietaygreen
I plugged in pi + 2pi(k) and got -1
jim_thompson5910
  • jim_thompson5910
did you first replace k with 0 and then simplify to get pi?
sammietaygreen
  • sammietaygreen
No. Oops. So I have to do cos(*insert equation*) and then have k be 0?
jim_thompson5910
  • jim_thompson5910
do you understand why I'm replacing k with 0?
sammietaygreen
  • sammietaygreen
bc it represents an integer?
jim_thompson5910
  • jim_thompson5910
yes, k is any integer and 0 is the easiest to work with because 0 makes things go away pi+2pi*k = pi+2pi*0 = pi + 0 = pi so `pi+2pi*k` is equivalent to `pi` when k = 0
jim_thompson5910
  • jim_thompson5910
you'll then find `cos(pi) = -1` so... cos(x) = -1 cos(pi) = -1 ... replace x with pi -1 = -1 ... true equation because the last equation is true, the first equation is true when x = pi so x = pi is definitely one of the infinitely many solutions x = pi is in the solution set for the equation cos(x) = -1
sammietaygreen
  • sammietaygreen
So how do I figure out which one is right?
jim_thompson5910
  • jim_thompson5910
do you agree that x = pi is a solution?
sammietaygreen
  • sammietaygreen
Well, that's one of the solutions I got as well so yes
jim_thompson5910
  • jim_thompson5910
ok try k = 1 and plug it into pi+2pi*k what is `pi+2pi*k` equivalent to when k = 1 ?
sammietaygreen
  • sammietaygreen
9.4 :/
jim_thompson5910
  • jim_thompson5910
in terms of pi
sammietaygreen
  • sammietaygreen
so it would be like 2*1 or 3.14 + 2*1?
jim_thompson5910
  • jim_thompson5910
pi+2pi*k = pi + 2pi*1 = ??? treat pi as if it were a variable
sammietaygreen
  • sammietaygreen
2pi?
jim_thompson5910
  • jim_thompson5910
think of it like 2 pies + 1 pie = how many pies?
sammietaygreen
  • sammietaygreen
3 pies lol
jim_thompson5910
  • jim_thompson5910
yep so 2pi+pi = 3pi
jim_thompson5910
  • jim_thompson5910
if k = 1, then pi+2pi*k = 3pi
jim_thompson5910
  • jim_thompson5910
tell me what cos(x) is equal to when x = 3pi
sammietaygreen
  • sammietaygreen
-1
jim_thompson5910
  • jim_thompson5910
now try k = 2. Plug it into pi+2pi*k to get what?
sammietaygreen
  • sammietaygreen
So I would just add the pi's the same way?
jim_thompson5910
  • jim_thompson5910
yeah it's similar to pi+2pi
sammietaygreen
  • sammietaygreen
so would it be 4pi?
jim_thompson5910
  • jim_thompson5910
pi+2pi*k = pi+2pi*2 = ?????
sammietaygreen
  • sammietaygreen
6? cause we're keeping it in terms of pi?
sammietaygreen
  • sammietaygreen
ill brb I have to workout
jim_thompson5910
  • jim_thompson5910
pi+2pi*k = pi+2pi*2 = pi+4pi = ?????
sammietaygreen
  • sammietaygreen
5pi?
sammietaygreen
  • sammietaygreen
cos(5pi) = -1 too
jim_thompson5910
  • jim_thompson5910
yep, let's do one more if k = 3, then pi+2pi*k is equal to what?
sammietaygreen
  • sammietaygreen
8pi?
jim_thompson5910
  • jim_thompson5910
just replace k with 3 and simplify \[\Large \pi+2\pi*\color{red}{k}=\pi+2\pi*\color{red}{3} = ???\]
sammietaygreen
  • sammietaygreen
wait, no 9
jim_thompson5910
  • jim_thompson5910
try again
sammietaygreen
  • sammietaygreen
wouldn't it be 9pi though? I don't see how it qwouldnt be

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