Question

Does anyone know how to solve this? csc^4(2X)-4=0 for all real numbers?

Answer 1

Hello tolli, first of all Welcome to OpenStudy.

Answer 2

First of all let me know that what you tried yet to solve this question ?

Answer 3

so, first I wrote csc in terms of 1/sin. so this is what I first wrote:1/sin^4x(2x)-4. Then I added the 4 to the other side: 1/sin^4(2x)=4. Then when I tried to factored/take out a greatest common factor in 1/sin^4x but it just got messy I I got lost.

Answer 4

OK! wait let me check your method

Answer 5

what I will prefer you to do is :
\[\large{\frac{1}{\sin^4(2x)} = 4}\]
\[\large{1 = 4\sin ^4 (2x)}\]
\[\large{\sqrt{1}= \sqrt{4\sin^4(2x)}}\]
What do you get now ?

Answer 6

+/-1=+/-2sin^2(2x)

Answer 7

OK leave \(\pm\) for once. Now ?

Answer 8

would I divide by 2 then square everything?

Answer 9

OK see here :
\[\large{1 = 2 \sin^2(2x)}\]
\[\large{\frac{1}{2} = \sin^2(2x)}\]

Right ?

Answer 10

Now, square root ...

Answer 11

Not square but square root

Answer 12

YES. then you square it to get: 1/4=sin(2x)

Answer 13

oh ok

Answer 14

square root it and then tell me what you get

Answer 15

sqrt2/2=sin(2x)

Answer 16

Correct, well done!. Or I can write that as :
\[\large{\frac{1}{\sqrt{2}} = \sin(2x)}\]

Answer 17

should I multiply sqrt2/2 by 2?

Answer 18

No. Not now

Answer 19

Can you tell me at which angle \(\large{\sin (x) }\) is \(\large{\frac{1}{\sqrt{2}}}\) ?

Answer 20

45

Answer 21

and 2pi/4

Answer 22

45 degrees is right but \(\large{\frac{2\pi}{4}}\) is not as 2pi/4 = pi/2 which is equal to 90 degrees not 45 degrees friend.

So you have 45 degree as pi/4 radians. ok?

Answer 23

ok

Answer 24

Now, we have : \[\large{\frac{1}{\sqrt{2}} = \sin(2x)}\]
We know that : \[\large{\sin (\frac{\pi}{4}) = \frac{1}{\sqrt{2}}}\]

right ?

Answer 25

so can you tell me what will be "x" now ?

Answer 26

pi/2

Answer 27

how ?

Answer 28

ok so, sin(pi/4)=1/srqt2, so if I double that I get 2pi/4 which simplifys to pi/2.

Answer 29

oh! You have a misconception there .

see : \[\large{\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}}\]
also \[\large{\sin (2x) = \frac{1}{\sqrt{2}}}\]

therefore : \[\large{\sin(\frac{\pi}{4}) = \sin(2x)}\] as they both are equal to the same thing

therefore pi/4 = 2x

Answer 30

can you tell me now , what is x equal to ?

Answer 31

can I take sin inverse of pi/4 then divide by 2?

Answer 32

See you can do like this :

\[\large{\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}}\]
\[\large{\frac{\pi}{4} = \sin^{-1}(\frac{1}{\sqrt{2}})}\]
now divide by 2

Answer 33

what happen to the sin(2x)?

Answer 34

now : \[\large{\sin(2x) = \frac{\pi}{4}}\]
\[\large{2x = \sin^{-1}(\frac{\pi}{4})}\]

now divide by 2 . can you do that now?

Answer 35

yes.but what is the sin inverse (pi/4)? Is it still pi/4?

Answer 36

oh sorry mistake there :

\[\large{\sin(2x) = \frac{1}{\sqrt{2}}}\]
\[\large{2x = \sin^{-1}(\frac{1}{\sqrt{2}})}\]

now divide by 2 , m sorry

Answer 37

so, x= 2/sqrt2=2sqrt2/2=sqrt2

Answer 38

no see now :

\[\large{\sin^{-1}(\frac{1}{\sqrt{2}}) = \frac{\pi}{4}}\]

therefore I have 2x = pi/4

x = pi/8

Answer 39

oh. so when writhing the equation because the original question asked for all real numbers, will it be:x=pi/8+2npi

Answer 40

npi +/- pi/8

Answer 41

why is npi instead of 2npi?

Answer 42

the formula for sin(pi/8) general soln is :

\[\large{n\pi + (-1)^n (\frac{\pi}{8})}\]

Answer 43

where is the -1^n coming from?

Answer 44

http://mathsfirst.massey.ac.nz/Trig/TrigGenSol.htm

Answer 45

I'm a bit disillusioned, here...
Might I suggest an alternative way to do it?
Starting from
\[\huge \sin^2(2x)=\frac{1}{2}\]

Answer 46

i would like that but I have to go to school. unless you can do it in 5min.

Answer 47

Okay, let's see :)
Then
\[\huge \sin(2x)=\pm\frac{1}{\sqrt{2}}\]

Answer 48

mathslovers thanks for all your help.

Answer 49

What are all the angles from 0 to 2pi that have sines of plus or minus 1/sqrt(2)
They are
\[\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\]

Answer 50

yes. I understand that

Answer 51

And these are possible values for 2x, so possible values for x are
halves of these...
\[\large \frac{\pi}{8},\frac{3\pi}{8},\frac{5\pi}{8},\frac{7\pi}{8}\]
plus half of 2pi, which is pi

Answer 52

so what is the general solution?

Answer 53

These differ by pi/4, so let's add pi to pi/8, we get 9pi/8, which is just 7pi/8 + pi/4
Hence, if you add any multiple of pi/4 to pi/8, you still get a solution...
The general solution is
\[\huge \frac{\pi}{8}+\frac{k\pi}{4} \ \ \ \ k \in\mathbb{Z}\]

Answer 54

That was a really quick summary, but that gets you there :)

Answer 55

Thankyou so much your method help a lot.