Question

medal!!
solve cos x tan x-sin^2x=0 for all real values of x

Answer 1

any ideas?

Answer 2

remember \[\tan x=\frac{\sin x}{\cos x}\]

Answer 3

yea..

Answer 4

this is really just algebra

Answer 5

so \[\cos x \frac{\sin x}{\cos x}=...\]

Answer 6

well \[a\frac{b}{a}=...\]
how would you do this

Answer 7

c?

Answer 8

no there is no c
how about \[\frac{ab}{a}=\]

Answer 9

b

Answer 10

yes its b! can you tell me why

Answer 11

because ab/a so u take the a's out and ur left with only b

Answer 12

yes you cancel a/a because a/a=1 and we know multiplying by 1 does nothing
\[a\frac{b}{a}\] this is the same thing you know heheh

Answer 13

okay now what to do with
\[\frac{\cos x \sin x}{\cos x}=...\]

Answer 14

sin x

Answer 15

yes
so we get \[\sin x-\sin^2 x=0\]
any idea now how to go from here

Answer 16

\(\large\color{slate}{ \displaystyle \cos x \tan x-\sin^2x=0 }\)
\(\large\color{slate}{ \displaystyle \sin x-\sin^2x=0 }\)
\(\large\color{slate}{ \displaystyle \sin x(1-\sin x)=0 }\)

Answer 17

again i would ask you how would you solve \[x-x^2=0\]

Answer 18

^2

Answer 19

slowly, indeed nice:)

Answer 20

you spoiled it solo haha

Answer 21

thanks though :)

Answer 22

oka can you tell me how can you solve \[x-x^2=0\]

Answer 23

x-x^2 so =^2

Answer 24

the same way will use to solve the other one

Answer 25

hmm no
\[x-x^2=0 \Longrightarrow x(1-x)=0 \Longrightarrow x=0~or~x=1\]

Answer 26

don't forget your algebra haha

Answer 27

ow lol kind of forgot about it

Answer 28

we gonna do the same thing:
\[\sin x-\sin^2 x=0 \Longrightarrow \sin x(1-\sin x)=0 \Longrightarrow \sin x=0 ~or \sin x=1\]

Answer 29

now we need to find what is x
\[\sin x=0~~or~~\sin x=1\]
what angles give us 0
and what angles give us 1

Answer 30

you know the unit circle right?

Answer 31

yea

Answer 32

so what angle give us 0 we we take sin of that angle?

Answer 33

sin (what)=0

Answer 34

rememeber this

Answer 35

yea

Answer 36

okay so you know sin(...)=0

Answer 37

tell what should i put in dots

Answer 38

0

Answer 39

yes sin(0)=0
now the question is
is that the only angle which has sin=0

Answer 40

so 90 too

Answer 41

here we solving for all x's so there are many other angles that has the same sin

Answer 42

oh no! look at the circle is that true

Answer 43

yea so whats next

Answer 44

but sin(90) isnt 0

Answer 45

pi/2 =90 look at the unit circle

Answer 46

okay i'm just gonna say it to you
sin(90)=1 not 0
=========
the other angles that give sin=0
are pi, 2pi, 3pi,4pi,5pi......
if we do rotation clock wise we have
-pi, -2pi, -3pi,-5pi... and so one
so we say any angle that is like this \[n\pi\] when n is an integer is a solution to \[\sin x=0\]
\[n=0,\pm1,\pm2,\pm3,\pm4.....\]

Answer 47

that's one part
now we ask what angles whose sin=1
we have already found one sin(90)=1
but again that's not the only one
every complete rotation would give us sin=1
so all multiples of 90 are solution to sinx=1
\[\sin x=1 \Longrightarrow x=\frac{\pi}{2}+2n\pi\] where \[n=0,1,2,3....\]

Answer 48

@cutegirl
i guess you don't get all this
you need real review of your algebra as well as the unit circle
pls watch some videos of this stuff

Answer 49

yea i will do that after this so whats the answer haha

Answer 50

and while i was helping you, you shouldn't go around that's not a good thing to do :(

Answer 51

check my comments i give the answer

Answer 52

yea, i had other question that why

Answer 53

well first do this and then do the other
jumping to other stuff is not helping

Answer 54

well i'm just saying... good luck

Answer 55

sorry my bad is it a because pi k, pi/2 +2 pi k

Answer 56

here is the answers: \[\sin x =0 \Longrightarrow x=n\pi, ~n\in \mathbb{Z}\]
\[\sin x=1 \Longrightarrow x=\frac{\pi}{2}+2n\pi, ~n\in \mathbb{Z^+}\]

Answer 57

so its not a?

Answer 58

it is a
i just made it n instead of k
i didn't see the choices you give me
but that does not matter anyway
n, k whatever as long as you understand what they stand for

Answer 59

thanks so much!!

Answer 60

welcome