Question

Problem: To solve this inequality without the aid of a calculator.

Answer 1

What inequality?

Answer 2

the question doesn't make any sense to me
you got a screenshot or sommat?

Answer 3

\[Cosx−sinx>0\]

Answer 4

\[\cos(x)-\sin(x)>0\]?

Answer 5

yup

Answer 6

#confused

Answer 7

there is a pretty snappy way to do it i think
do you know how to write that as a single function of sine or cosine ?

Answer 8

:/ i did it the only way i know?

Answer 9

your way doesn't look terrible

Answer 10

i did it in tangent, but that in itself isnt easy

Answer 11

but is has a mistake in it i think

Answer 12

how do you go from \[\cos(x)>\sin(x)\] to \[\tan(x)<1\]?

Answer 13

if we divide both sides by cos x , we have to consider the possibility that cos x is positive or negative..right?

Answer 14

oh. i divided both sides by sin(x)

Answer 15

or i guess i should say "you cannot" since you have an inequality, you can't divide both sides by an expression if you don't know if it is positive or negative

Answer 16

hmm...okay then how do i do it?

Answer 17

how? how did you do that?

Answer 18

maybe you have not seen this, in which case i am stumped as to how to continue
see if you your book it says how to write \[a\sin(x)+b\cos(x)\] as \[\sqrt{a^2+b^2}\sin(x+\theta)\][

Answer 19

where \[\tan(\theta)=\frac{b}{a}\]

Answer 20

in your case you have \[-\sin(x)+\cos(x)=\sqrt{1^2+1^2}\sin(x+\theta)\] where \[\tan(\theta)=-1\]

Answer 21

im in online school...and it doesnt make much sense at all

Answer 22

oh sorry

Answer 23

well we can continue anyways if you like

Answer 24

I thought we didnt have a tangent though

Answer 25

since \[\tan(\theta)=-1\] you can put \[\theta=\frac{3\pi}{4}\] and use \[\sqrt{2}\sin(x+\frac{3\pi}{4})\]

Answer 26

we don't in the problem, i just rewrite the left hand side of the inequality as \[\sqrt2\sin(x+\frac{3\pi}{4})\]

Answer 27

oh. okay

Answer 28

but then this is wrong?
http://www.wolframalpha.com/input/?i=solve+cos+x+-+sin+x+%3E+0

Answer 29

how do i get that answer?

Answer 30

no wolfram is never wrong
sure you would get the same thing

Answer 31

oh except i am an idiot, you want postiive, i solved for negative sorry

Answer 32

\[x+\frac{3\pi}{4}=0\] solve that

Answer 33

lol you arent an idiot. i didnt know any of this stuff ll

Answer 34

and also solve \[x+\frac{3\pi}{4}=\pi\]
that will give you one interval

Answer 35

- 246/100?

Answer 36

and

Answer 37

first one gives \[x=-\frac{3\pi}{4}\] second one gives \[x=\frac{\pi}{4}\]

Answer 38

oh you meant like that lol

Answer 39

yeah so \[-\frac{3\pi}{4}<x<\frac{\pi}{4}\] is one interval

Answer 40

and then how does each side have \[2\pi~n\]

Answer 41

and generalize by adding \(2n\pi\) to each sides since sine is periodic with period \(2\pi\)
that explains wolframs answer \[2n\pi-\frac{3\pi}{4}<x<2n\pi+\frac{\pi}{4}\]

Answer 42

OHHHHHH

Answer 43

lol

Answer 44

can you delete the ones for negative? that way I can save this question and review it later for my test?

Answer 45

(and understand it?)

Answer 46

oh sure i get it hold the phone

Answer 47

thanks :)

Answer 48

did it
all steps are there, not sure if you missed anything but this was the key
you can always write \[a\sin(x)+b\cos(x)\] as a single function of sine \[\sqrt{a^2+b^2}\sin(x+\theta)\] where \[\tan(\theta)=\frac{b}{a}\] that is how you find the \(\theta\) in that formula

Answer 49

yw

Answer 50

:D thnx

Answer 51

@RhondaSommer did u copy from nightowl?

Answer 52

no. I had the same problem. I gave nightowl her explanation too :D it really makes sense

Answer 53

okay
good for u <3

Answer 54

:D I even gave her credit...sorta. should I write her name? that would be better right?

Answer 55

i did anyway

Answer 56

no dear
i confirmed whether u understood or not

Answer 57

oh. yes i understand :P I figured it helped me so it might help night owl