Question

List the solutions:
A) sin x -1 = cosx
B) sin(2x)-1=tanx

Answer 1

I think the solution to A) is x=pi/2, x=pi

Answer 2

a) sin(x) - 1 = cos(x),
did I copy correctly?

Answer 3

is that the question

Answer 4

I'm not sure. It just says sin x - 1= coax, but I think you did

Answer 5

how about squaring both sides

Answer 6

could you show me? Are my answers to A) wrong?

Answer 7

@perl

Answer 8

(sin x -1 )^2 = (cos x) ^2
square both sides we get
sin^2 (x) - 2*sin(x) + 1 = cos^2 (x)

Answer 9

ok so far?

Answer 10

I believe so

Answer 11

square both sides gives us
(sin x - 1 )(sin x - 1) = cos x * cos x
then i foiled the left side

Answer 12

so we have
sin^2 (x) - 2*sin(x) + 1 = cos^2 (x)
now on the right side, cos^2(x) , we can substitute 1 - sin^2(x) for it

Answer 13

Here are the steps again:
sin x - 1 = cos x
(sin x - 1 )^2 = (cos x)^2
sin^2 (x) - 2*sin(x) + 1 = cos^2 (x)
sin^2 (x) - 2*sin(x) + 1 = 1 - sin ^2 (x)

Answer 14

Ok I get that so far

Answer 15

we can substitute 1 - sin^2(x) for cos^2(x) since they are equal. do you see why they are equal?
because sin^2 (x) + cos^2(x) = 1 , and subtract sin^2(x) from both sides

Answer 16

cos^2(x) = 1 - sin^2(x) now is a new identity

Answer 17

ok, so whats the next step @perl

Answer 18

Ok I'm following so far @perl now we need to solve for x i think

Answer 19

sin x - 1 = cos x
(sin x - 1)^2 = (cos x)^2
sin^2 (x) - 2*sin(x) + 1 = cos^2 (x)
sin^2 (x) - 2*sin(x) + 1 = 1 - sin ^2 (x)
2sin^2(x) - 2 sin(x) + 1 = 1
2 sin^2(x) - 2sin(x) = 0
2sin(x)*( sin (x) - 1) = 0

using zero product property
2 sin (x) = 0 OR sin(x) - 1 = 0
sin(x) =0
x = pi*n

sin(x) -1 = 0
sin(x) = 1

x = pi/2 + 2pi*n

Answer 20

oh so I was right lol?

Answer 21

what is the domain , between 0 and 2pi ?

Answer 22

yes

Answer 23

actually the general solution is
x = pi + 2pi*n, pi/2 + 2pi*n

apparently it is not x = pi*n, since if n = 0 , x= pi*0 = 0 is a false solution

Answer 24

oh ok so i;ll use the general solution answer. what about b)? I have to leave soon so please help like you just did @perl

Answer 25

one second

Answer 26

this one is tricky

Answer 27

Yes I know. This is the one I really needed help on

Answer 28

this one you have to use a calculator

Answer 29

graph the two functions separately and estimate where they intersect

Answer 30

use the zero option

Answer 31

Could you give me the solution and show me how you got there? I really have no clue

Answer 32

walk me through the steps would be great if you can. It's just I have to leave soon

Answer 33

graph
y = sin(2x)-1
y= tan(x)

Answer 34

graph them separately,

Answer 35

I have the graph, but it doesn't show me the solutions. could you tell me?

Answer 36

sure

Answer 37

do the directions say what the domain is ?

Answer 38

should i include the 2pi*n on both parts? and yes 0,2pi

Answer 39

if the problem says between 0 and 2pi, then don't include 2pi*n

Answer 40

but you need to calculate
2pi ~ 6.28
so you want solutions between 0 and 6.28

Answer 41

wait I'm confused now. I thought the answers were:
A) x=pi/2, pi
B)x= -0.59, x= 1.68

Answer 42

the domain is [0,2pi] for both

Answer 43

for a) x = pi/2, pi
b) x = 2.0687813 , 5.210374

Answer 44

oh ok. thanks for helping me here. Hopefully your correct :)

Answer 45

does it say how much accuracy?

Answer 46

no, just find the solutions

Answer 47

I checked maple, by the way

Answer 48

wait, again. what other the final solutions for B) . I see so many different answers

Answer 49

oh shoot, that was a mistake again

Answer 50

@perl so what is the final answer lol?

Answer 51

\[\sin x-\cos x=1,squaring~ both~ sides\]
\[\sin ^2x+\cos ^2x-2\sin x \cos x=1\]
1-sin 2x=1,
sin 2x=0\[\sin 2x=0=\sin n \pi ,2x=n \pi,x=\frac{ n \pi }{ 2 }\]
where n is an integer.

Answer 52

why isn't this showing up for me?

Answer 53

oh and @surjithayer I already know the answer to the first one. I need the 2nd

Answer 54

sorry the program crashed

Answer 55

@perl that;s ok. did you figure it out?

Answer 56

a) x = pi/2, pi
b) x = 2.0687813 , 5.210374

Answer 57

There's no exact solution to the second answer -- it'll be an irrational number. An approximate solution, of course, is possible. However, you seem to be confused by the notion that there are an infinite number of solutions, but the only solutions we are interested in is when they fall within some domain, namely, in this case, [0, 2pi].

Answer 58

Well, an irrational number that isn't a multiple of pi, and such, that is. :)

Answer 59

Ok thanks guys. I believe @perl is correct for B) :)

Answer 60

Perl's answer is correct.

Answer 61

here, I put it into wolfram
http://www.wolframalpha.com/input/?i=solve+sin%282x%29-1%3D+tan%28x%29+%2C+0%3Cx%3C2pi

Answer 62

Just go to WolframAlpha and type: sin(2x)-1=tan(x).

Answer 63

Yup.

Answer 64

And look for solutions within the desired interval, e.g., sin(2x)-1=tan(x), from x=0 to x=2pi

Answer 65

type:
solve sin(2x)-1= tan(x) , 0<x<2pi

Answer 66

queelius, wolf gives me an ugly answer if i dont specify x is between 0 and 2pi

Answer 67

queelius, compare to this
http://www.wolframalpha.com/input/?i=solve+sin%282x%29-1%3D+tan%28x%29+

Answer 68

wolfram general solution is :
x = 2 tan^(-1)(root of x^6+2 x^5+x^4-12 x^3-x^2+2 x-1 near x = -0.594554)+2pi *n

x= 2 tan^(-1)(root of x^6+2 x^5+x^4-12 x^3-x^2+2 x-1 near x = 1.68193)+2pi*n

Answer 69

Perl, lol, that's when you ask for an exact answer. It gives you a nice formula dependent upon n, n any integer, but as you indicated, you need to choose an n for which the result falls within the desired interval. You can also just hover over the intersection points on the plot to get an approximate answer.

Answer 70

thats a clue that there is no simple solution , in terms of pi or such

Answer 71

what is interesting in part a) , when we squared both sides of the equation we introduced an extraneous solution

Answer 72

ok guys lol is perls answer correct?

Answer 73

yes wolfram concurs with me, and i did it using a TI 84, graphed and intersected

Answer 74

ok thanks

Answer 75

Squaring both sides does allow for this: (-1)^2=(1)^2,

Answer 76

So you have to filter out extraneous solutions afterwards.

Answer 77

Btw, perl, earlier when I said "you seem to be confused by the notion that there are an infinute number..." I wasn't directing that to you, but to washcaps.

Answer 78

I just realized you may have thought I was referring to you. Random aside. :)