Question

PLs Help.......Medal..

Answer 1

I have no idea. Sorry.... @cambrige

Answer 2

@Data_LG2

Answer 3

thanks anyways

Answer 4

@terenzreignz

Answer 5

@Orion1213

Answer 6

Just sub in each of the four values and see if it satisfies the equation?

Answer 7

went thru my mind...But I need the solution process!!

Answer 8

@PixieDust1 @kawaiicat123

Answer 9

@ikram002p

Answer 10

sarahusher's suggestion is the most straightforward, although I'd suppose it's possible to solve the given inequality for x.

If we do substitute x = Pi/6, we get Abs(Sqrt(3) * Sqrt(3)/2 - 1/2), or Abs(3/2 - 1/2). Please judge for yourself whether x = Pi/6 is a solution or not.

Answer 11

I'll give you a hint, only one of them satisfies it

Answer 12

lol

Answer 13

Its an MCQ.I think I got that

Answer 14

yeah, i think that's a better way (^_^)
i tried solving it but i got stuck.. http://sketchtoy.com/56878404
lol, sorry..

Answer 15

no substitutions...........Either graph or via formula

Answer 16

good try @Data_LG2 appreciae it!!:)

Answer 17

Cos(x) can only be anywhere from 1 to -1. Sin(x) has the same range. sqrt(3)cos(x) can only be from sqrt(3) to -sqrt(3), which is about 1.73 to -1.73.

Answer 18

yup

Answer 19

substitution does give me A

Answer 20

How come you can only use graph or formula?

Answer 21

not graph.........what I want is a method where I don't have to substitute all the options!!

Answer 22

@experimentX @emilyhaddad @oksuz_ @ash2326

Answer 23

So knowing that, now we need to find when the difference between sqrt(3)cos(x) and sin(x) is greater than 2. How did you get A as the answer?

Answer 24

I got sqrt(3)cos(pi/6)-sin(pi/6)=1

Answer 25

(sorry you said graph above)
umm I'm not using a formula would be particularly effective, as the interval of [0, 4pi] can be infinitely tested.
If you are adamant on using a formula, I can't help. Sorry.

Answer 26

A is incorrect

Answer 27

And you can probably set up some calculus equation to solve for every x that makes this true.

Answer 28

no I did not I got A for <=2.....lol not >

Answer 29

-_-

Answer 30

As said, it's pointless working a formula for a multiple choice question that can take two minutes, so I'll leave you guys to it @Snowfire looks like she's got it!! :-)

Answer 31

Oh no, I don't xD I just know what the formula will be based on, nothing else atm

Answer 32

Similar to related rates problems

Answer 33

divide and multiply the mod term by rt 2.... it'll become rt 2 | rt 3/rt2 cos -1/rt2 sin x|
combine the whole term by using cos (A-B) or Sin (A-B) as sin pi/4 = 1/rt2
Rest is simple by mod

Answer 34

cos (A+B)**

Answer 35

you can use

cos(x+y)=cosx cosy-sinx siny

Answer 36

dividing both sides by 2, we'll have an obvious relationship with pi/6...

Answer 37

can u write the formula @divu.mkr @Orion1213

Answer 38

you'll get x <= 11pi/6, take note that x ranging from 0 to 4pi

Answer 39

I meant the method

Answer 40

\[\left|\frac 12 \sqrt3 (e^{-i x}+e^{i x})-\frac12 i (e^{-i x}-e^{i x})\right|\ge 2\]
\[\left|\sqrt3 (e^{-i x}+e^{i x})- i (e^{-i x}-e^{i x})\right|\ge 4\]
dunno how useful that would be tho

Answer 41

For anyone interested in exploring this further:
Start by assuming that the quantity to which the absolute value operator is applied is already positive or zero. Then the absolute value op can be omitted. This leaves us with Sqrt(3)*cos x - sin x (= to or greaterhan t) 2.
This can be re-written as Sqrt(3) cos x (= to or greater than) 2 + sin x.
Graph both Sqrt(3) cos x and 2 + sin(x) simultaneously on the same calculator screen, from 0 to 4Pi.
You'll see that Sqrt(3)*cox x is never greater than 2+sin x.
However, Sqrt(3)*cos x = 2 + sin x for (what appears to be) two different x values.
It's possible to use the TRACE function on the calculator to determine the x-coordinate of the point (or points) of intersection.
And so on.

Answer 42

\[\left|\sqrt{3}\cos x-\sin x\right|\ge2\]\[\left|\frac{\sqrt{3}}{2}\cos x-\frac{1}{2}\sin x\right|\le1\]\[\left|\cos(\frac{\pi}{6}+x)\right|\le1\]\[\left|\frac{\pi}{6}+x\right|\le\cos^{-1}(1)\]\[\left|\frac{\pi}{6}+x\right|\le2\pi\]\[\left|x\right|\le2\pi-\frac{\pi}{6}\]\[\left|x\right|\le\frac{11\pi}{6}\]

Answer 43

\[\left|x\right|\le-\frac{\pi}{6}\]

Answer 44

Really, really cool approach, Orion! Congrats.

Answer 45

Thanks... :)

Answer 46

hey @Orion1213 how did u get <=pi/6??

Answer 47

and how did u change the inequality???

Answer 48

remember when we do division or multiplication in inequality, the inequality revert in favor of the other...

Answer 49

and for pi/6, from the unit circle as well as identity cos(x+y)=cosy cosx - siny sinx... same as the format of the problem...
cos y = sqrt(3)/2
sin y = 1/2
it comes out y=pi/6

Answer 50

in ur formula,why did u change the inequality??

Answer 51

sry.how cud u change ur inequality>??

Answer 52

i cross multiply 2 remember, that means i divide both sides by 2 so there will be a change in inequality...

Answer 53

u did not multiple with a -ve sign!!

Answer 54

i only use -ge then -le sign

Answer 55

inequality changes only when u divide or multiply with a -ve sign!!

Answer 56

sorry i can't get what you mean by -ve sign, there are only two signs i have used and those are -le (for less than or equal to) and -ge (for greater than or equal to) signs...

Answer 57

-ve means negative
+ve means positive

Answer 58

oh i see... thanks for letting me know... :)

Answer 59

ok i found my mistake, i should have not change the inequality sign unless i multiplied or divide by -ve sign, i reviewed it... so the final inequality must be \[\left|x\right|\ge-\frac{\pi}{6}\]thanks to all, I've learned a lot. It is only now I observe THAT RULE. Thanks @cambrige.... :)

Answer 60

Ur Welcome...Thank U too...for the effort u put!!