Question

how do you do cos2x=0, the answer says 3pi/4

Answer 1

Depends if you're integrating or derivating

Answer 2

oh, sorry, derivating.

Answer 3

cos(2x) = 0
2x = arcos(0)
2x = 3pi / 2
x = 3pi/4

Answer 4

\[\frac{ d }{ dx }\cos2x\]

Answer 5

what is arcos?

Answer 6

Yeah...DD really knows his unit circle.

Answer 7

arc cos is the inverse cosine used to find the angle of measurement

Answer 8

cos equals o at pi/2 and 3pi/2?

Answer 9

arc cos = \[\cos ^{-1}(\theta)\]

Answer 10

arccos is the inverse cosine function:
arccos(0) is the values of theta where cos(theta) = 0

Answer 11

yes, but cos equals 0 at pi/2 and 3pi/2 so how do you know which one to use?

Answer 12

In general, when :

\( cos(x) = 0\)
Then:

\[x = (2n+1) \frac{\pi}{2}\] where n belongs to set of Integers..

Answer 13

It depends. Usually the question will include a range of possible theta values, either from 0 to pi, or 0 to 2pi, or something. If they don't give you the range, either value should work, even if it's not the answer in the book.

Answer 14

You'd know which one to use depending on the angle of measurement you're looking for.

Answer 15

Or which radians you're given and are working towards finding the angle measurement

Answer 16

what if i was finding a minimum value from an equation? would i just list both?

Answer 17

Im not sure what you mean by minimum value?

Answer 18

i don't understand how DDCamp got from..
2x = arcos(0)
to
2x = 3pi / 2
does arcos(0) just mean where cos is equal to 0?
we haven't covered ar? what does it mean?

Answer 19

I'mout, @genius12 , @waterineyes or @DDCamp will helpyou out with your problems! :D Gooooood luck!!!

Answer 20

Yes, arccos(0) is where cos(theta)=0

Answer 21

Who took your wicket @Jhannybean or you are Retired Hurt??? :P

Answer 22

Yes, DDcamp should explain why he chose to use \[\frac{ 3\pi }{ 2 }\] instead of \[\frac{ \pi }{ 2 }\] to solve for x

Answer 23

lol @waterineyes , I have to go study for a final!!! Engllish...=_=

Answer 24

From what it looks like, I believe you're solving for x. And I'm also assuming that your solutions are to exist in the interval [0, 2pi]. If the aforementioned conditions hold, then this is how you would solve for x:\[\bf \cos(2x)=0\]The only places where Cosine is 0 in the given interval is at pi/2 and 3pi/2, which means:\[\bf \implies 2x=\frac{\pi}{2} \ or \ 2x=\frac{3 \pi}{2} \implies x=\frac{\pi}{4},\frac{3 \pi}{4}\]
Do you understand?

Answer 25

Hey use the above one that I gave, put n = 1 there..

You can put 0, 1, 2 or -1, -2 all but according to your answer n = 1 there..

Answer 26

@genius12 found the write explanation for this problem :)

Answer 27

\[\cos(2x) = 0\]
\[\implies 2x = (2n+1)\frac{\pi}{2}\]

Put n= 1 here :

\[2x = \frac{3\pi}{2}\]
\[x = \frac{3\pi}{4}\]

Answer 28

okay, i understand, thankyou all!

Answer 29

how do you solve y=sin(2x+pi/4) to find the tangent and normal when x=0