Question

Solve the equation exactly in radians: cos^2x=1/2sin(2x), all x

Answer 1

\(\large\color{black}{ \displaystyle \cos^2x=\frac{1 }{2\sin(2x)} }\) like this?

Answer 2

sin(2x) is in the numerator

Answer 3

\(\large\color{black}{ \displaystyle \cos^2x=\frac{\sin(2x) }{2} }\) right?

Answer 4

\[\cos ^{2}x=(1/2)\sin(2x)\]

Answer 5

yes, I got it then, it is the same exact thing

Answer 6

There is a rule that: \(\large\color{brown}{ \displaystyle \sin(2x)=2\sin(x)\cos(x) }\)
first, rewrite it like this

Answer 7

\(\large\color{black}{ \displaystyle \cos^2x=\frac{2\sin x \cos x }{2} }\)

Answer 8

the 2's cancel, and you get: \(\large\color{black}{ \displaystyle \cos^2x=\sin x \cos x }\)

Answer 9

can you finish from here?

Answer 10

(subtract sin(x)cos(x) from both sides and factor)

Answer 11

I understand so far & no:(

Answer 12

I can factor out a cosx right?

Answer 13

\(\large\color{black}{ \displaystyle \cos^2x=\sin x \cos x }\)

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

yes, factor out of cos(x)

Answer 14

so i should get cosX(cosX-sinX)=0

Answer 15

yes, correct

Answer 16

then, either cos(x)=0 OR cos(x)-sin(x)=0

Answer 17

solve each of them

Answer 18

for cos(x)=0 i got \[\pi/2\] & \[3\pi/2\]

Answer 19

wait, your interval is then supposed to be \([0,2\pi]\) ?

Answer 20

are you looking for solutions over this (or some other) interval, or a general solution?

Answer 21

it says "all X"

Answer 22

then you need all possible answers

Answer 23

For \(\cos(x)=0\) , your solutions are:
\(\pi/2\),
\(\pi/2~~+\pi\),
\(\pi/2~~+2\pi\),
\(\pi/2~~+3\pi\),
and so on....


right?

Answer 24

and so would be true if you subtract Pi's instead of adding them

Answer 25

So,

\(\LARGE x={\rm n}\cdot\pi+\frac{1}{2};~~~\left\{n\in {\bf Z}\right\}\)

Answer 26

the last { } tell you that "n" can be any integer

Answer 27

okay and cosX-sinX= 0, how do you do that one

Answer 28

\(\large\color{black}{ \displaystyle \cos(x)-\sin(x)=0 }\)

\(\large\color{black}{ \displaystyle \cos(x)=\sin(x) }\)

do you remember the angle where sine and cosine functions are equal ?

Answer 29

pi/4?

Answer 30

yup, there you go

Answer 31

can you give me some other examples?

Answer 32

5pi/4?

Answer 33

yes, (you are adding pi), very good

Answer 34

or, you can have -3pi/4 , can you?

Answer 35

So you can go on subtracting \(\pi\) or adding \(\pi\), right ?

Answer 36

Yes but how do you know the equation to write? it always confuses me like the solution for cosX=0 and you have multiple solutions. if that makes sense

Answer 37

that is adding \(n\cdot \pi\) (where when n is a negative integer you are subtracting)

Answer 38

you have multiply solutions because you can go around many times (if you imagine a unit circle)

Answer 39

Our teacher uses 'K' and just says which K is any real #

Answer 40

these are the "coterminal angles"

Answer 41

K instead of n, then we can write K

Answer 42

capital or lowercase k ?

Answer 43

Capital

Answer 44

Okay, please try to write the solution set for \(\large\color{black}{ \displaystyle \cos(x)=\sin(x) }\)

Answer 45

(I will correct if anything)

Answer 46

well, what is the pattern, can you tell me?

Answer 47

adding pi?

Answer 48

yes, but you are adding any number of pi, you are adding 1pi , 2pi, 3pi, and so on..... OR, you are subtracting 1pi, 2pi, 3pi, (which is, adding -1pi, -2pi, -3pi ... and on)

\(\large\color{black}{ \displaystyle \frac{ \pi }{4} }\)
this is our starting point, and then you are adding these negative or positive pi's

\(\large\color{black}{ \displaystyle \frac{ \pi }{4} +K\pi }\)

(K can be negative, because it is an integer; this pattern perfectly satisfies the solution to the equation)

Answer 49

And then add a notation that K in an integer

\(\large\color{black}{ \displaystyle \frac{ \pi }{4} +K\pi;~~n\in{\bf Z} }\)

Answer 50

If anything seems not very clear, please ask....

Answer 51

Ok understood and the first one (cosX=0) how does the equation look with K?

Answer 52

I thought I have already given it, ... I will go through that one anyway

Answer 53

your starting point is Pi/2
then you add or subtract K\(\pi\) from that (where K is an integer)

Answer 54

without looking back into the thread, can you write the pattern for me please?

Answer 55

\[\pi/2+K \pi \]?

Answer 56

yes

Answer 57

where K is an integer