Question

Find, without using a calculator, all the possible values of y between 0 and 2pie for which tan(2y + pie/3 ) = 1

Answer 1

period of tan is pi

Answer 2

\[\tan(2y+\frac{\pi}{3})=1====>2y+\frac{\pi}{3}=\pi/4 +k \pi\]
with k in {.....-1,0,1,2.....}
find k wich satisfies y in 0 2pi

Answer 3

Erm just to verify do you get y = - pie/ 24 , 11pie/ 24 , 23pie/24 , 35pie / 24

Answer 4

the first one is not incude ! -pi/24<0 :) !

Answer 5

I beg your pardon?

Answer 6

we are looking for y beetween 0 and 2pie so ! not all values of k are good ! you have to choose ! which ones satifies y beetween 0 and 2 pi ! Did you understand ?!

Answer 7

yes i did the which quadrant method and tan y is in the first quadrant thus, basic angle pie/ 4.

Answer 8

tan y = 1*

Answer 9

so the answers are ...?!

Answer 10

i already typed my answers duhhhh above.

Answer 11

what answer :) ?!

Answer 12

y = - pie/ 24 , 11pie/ 24 , 23pie/24 , 35pie / 24

Answer 13

A=2y and B=pie/3

tan (A + B) = (tan A + tan B)/(1 - (tan A)(tan B)) now put the values of A and B

Answer 14

i hope its easy for you.

Answer 15

y=-pi/24 don't satisfy the condition y>=0 see ?@Sgstudent

Answer 16

and you missed one ! @Sgstudent

Answer 17

are you there ? @Sgstudent

Answer 18

i am trying muhammad method.

Answer 19

to be honest this is my first time attempting tan (A+B) i have attempted sin and cos (A+B) Oh welll

Answer 20

how to do his method lol

Answer 21

as in i need the workings.

Answer 22

wait

Answer 23

:D :D :D ! what you didn't understand in ""my method" ?!
and also muhammad's methods suppose that 2y==/==pi/2 and tan(2y).tan(pi/3) ==/==1
so you're solving an equation ! just it's inappriopriate to do hypothisis for the unkown :) !
I can show a full answer if you want to ! (there are two )

Answer 24

anything :)

Answer 25

i think A+B is easier

Answer 26

i love solving eqn :) :DD

Answer 27

I don't think so :) ! A+B is more dangerous !

Answer 28

my brain is different oh yea

Answer 29

did you understand the first part ?

Answer 30

no

Answer 31

lol

Answer 32

i dont know how to get tan 2 y values

Answer 33

let cange the problem ! Can you solve the equation
\[\tan(\alpha)=1\]

Answer 34

no

Answer 35

x= tan inverse

Answer 36

that's strange ! you said that you love solving equations ! :)

Answer 37

x = tan inverse wut/

Answer 38

kk I'll make you understand :) wait a minute

Answer 39

i just solved a tedious quadratic equation haha :P

Answer 40

look at the attachement ! to understand tan(x)=1 first !

Answer 41

quadratic equations are really boring :) !

Answer 42

done understanding

Answer 43

but trig equations are fascinating ! (sometimes not always !)
did you understand ! ?

Answer 44

i understandddddddddddddddddddd

Answer 45

are you angry ?! or what !
now do you agree with tan(smthing)=1<=====>smthing=pi/4+ k pi

Answer 46

i dont have anger management prob just i understand already :P

Answer 47

we continue ?

Answer 48

yes sure "_"

Answer 49

what does this face mean ?!
we said that tan (smthing)=1<=====>smthing=pi/4+k pi (wich you understood certainly) !
\[\tan(2y+\frac{\pi}{3})=1<=====>2y+1= \frac{\pi}{4}+k \pi \]
with k is an integer (this part is cleaar )

Answer 50

this face means that i am happy to learn more

Answer 51

:D :D you're making me happy ! then !

Answer 52

is it clear shall we continue ?

Answer 53

yes yes keep it going

Answer 54

\[2y=-\frac{\pi}{3}+\frac{\pi}{4}+k \pi =- \frac{\pi}{12}+ k \pi =\frac{12k \pi-\pi}{12}\]
then \[y=\frac{12k \pi -\pi }{24}\]
is this part cleaar .? ( now I need all your mind )

Answer 55

FOCUS*

Answer 56

now we will discuss about this kk ?§

Answer 57

OK

Answer 58

now they want us to find y beetween 0 an 2pi
now we have to choose k ! if we choose k=-1 (for example) is " y valid ?! is y beetween 0 an 2pi !

Answer 59

please don't leaave me alone ! :):)

Answer 60

I am here :(

Answer 61

you are not alone , i am here with you , though you far away i am here to stay lol quoted by MJ

Answer 62

ohh ! I was checking your profile :) !
now :) Answer the question !

Answer 63

What was the question?

Answer 64

now they want us to find y beetween 0 an 2pi now we have to choose k ! if we choose k=-1 (for example) is " y valid ?! is y beetween 0 an 2pi !

Answer 65

it is unvalid because it cant be <0 theeeeeeeeee

Answer 66

are you angry again ? now k=0

Answer 67

i am not angry i am very happy LOLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLLL

Answer 68

k = 0 also cannot :(

Answer 69

k=1 ; k=3; k=4 ;k=5,k=6( now ! I bet you're not happy )

Answer 70

k=2 :D don't forget

Answer 71

nonsense LOL

Answer 72

the second method a lot easier
\[ 0 \le y \le 2 \pi \Rightarrow 0 \le \frac{12k \pi -\pi}{24} \le 2 \pi\]
which Implies \[0 \le \frac{12k-1}{24} \le 2 \]
then \[1 \le 12 k \le 49\]
\[\frac{1}{12}\le k \le \frac{49}{12}\]
because k is an integer k =0 or 1 or 2 or 3 or 4
then you'll have 5 solutions ! I hope it's clear now !

Answer 73

are you there Hapyy man or woman ?!