Question

solve on the interval of [0,2pi)
2sin^2 3sinx-1=0

Answer 1

so far I tried using the quadratic equation but I got numbers that i dont understand how to deal with. Please help!

Answer 2

You are correct on using the quadratic formula, but before you must do that
You must substitute something else for "Sin"

Answer 3

we have a cubic unless there is a type-o in the question

Answer 4

well I plugged in the numbers as
a=2 b=3 and c=-1

Answer 5

nevermind it isn't a cubic
I see \[2\sin^2(3 \sin(x))-1=0\]
is that right?
I thought I see a cubic because I thought I read:
\[2\sin^2(x) 3\sin(x)-1 =0 \text{ which is just } 6 \sin^3(x)-1=0\]

Answer 6

okay so that equation can simplify to 6sin^3(x)-1=0??

Answer 7

if you meant:
\[2 \sin^2(x)3\sin(x)-1=0?\]

Answer 8

but you wrote
\[2 \sin^2(3 \sin(x))-1=0\]

Answer 9

yes i was wrong the x was supposed to be included

Answer 10

so now what? since its cubic how do i solve it

Answer 11

so the equation really is
2 sin^2(x)3sin(x)-1=0?

Answer 12

yes

Answer 13

are you you seem to have thought the question was 2sin^2(x)+3sin(x)-1=0 earlier but I'm not sure what you mean the question to be

Answer 14

i am just wondering how to solve this equation

Answer 15

lol I'm very confused what you mean the question to be
your solution seems to suggest you wrote the question here wrong

Answer 16

you guys called it a quadratic where you picked a=2 and b=3 and c=-1
this seems to suggest the equation is actually 2sin^2(x)+3sin(x)-1=0

but you are saying the equation is 2sin^2(x)3sin(x)-1=0

just want to be sure

Answer 17

The equation is 2sin^2(x)3sin(x)-1
How do I solve this?
I tried using the quadratic equation but you added that it is a cubic equation so it cannot e solved using the quadratic equation. Now what? How is this solved then

Answer 18

do you know 2 times 3 is 6?
and sin^2(x) times sin(x) is sin^3(x)?

Answer 19

\[2 \sin^2(x) 3 \sin(x)-1=0 \\ 2 \cdot 3 \sin^2(x) \sin(x)-1=0 \\ 6 \sin^3(x)=1 \\ \sin^3(x)=\frac{1}{6}\]

Answer 20

Im confused because you posted the same equation multiple times asking me if thats thrright one but you are asking me about two of the same exact equations huh

Answer 21

I really think you meant this equation though:
\[2\sin^2(x)+3\sin(x)-1=0\]

Answer 22

Okay so that is the eqaution simplified and solved?

Answer 23

what?

Answer 24

2sin^2(x)3sin(x)-1=0 is totally not the same as 2sin^2(x)+3sin(x)-1=0

Answer 25

I want you to tell me the equation
I don't want to guess anymore what the equation is

Answer 26

Ohh I see! Yeah so there is no addition sign inbetween
the equation is 2sin^2(x)3sin(x)-1=0

Answer 27

therefore what you did when you got the product sin^3(x)=1/6 is correct I think

Answer 28

Sorry for making this confusing

Answer 29

yes if you really meant the equation is 2sin^2(x)3sin(x)-1=0

now take the cube root of both sides

Answer 30

okay

Answer 31

I get sinx=.55

Answer 32

shoot did i just delete those?

Answer 33

\[\sin(x)=\sqrt[3]{\frac{1}{6}} \\ \text{ recall } \sin(x) \text{ has period } 2 \pi \\ \sin(x+2 \pi n)=\sqrt[3]{\frac{1}{6}} \\ \text{ now do } \arcsin( ) \text{ on both sides }\]

ok also recall
\[\sin(-x+\pi)=\sin(x) \text{ since } \\ \sin(-x) \cos(\pi)+\sin(\pi) \cos(-x) \\ -\sin(x)(-1)+0 \\ \sin(x) \\ \text{ so remember we period } 2\pi \\ \text{ so } \\ \sin(x)=\sin(-x+\pi +2 n \pi )=\sin(-x+\pi(1+2n)) \\ \text{ so you also have to solve } \\ \sin(-x+\pi(1+2n))=\sqrt[3]{\frac{1}{6}}\]
but you can do this an s similar way
take arcsin( ) of both sides

Answer 34

so basically you have to solve
\[\sin(x+2 n \pi)=\sqrt[3]{\frac{1}{6}} \text{ and } \sin(-x+\pi(1+2n))=\sqrt[3]{\frac{1}{6}} \\ \text{ using that \arcsin function } \\ \\ \text{ you have } \\ x+2n \pi=\arcsin(\sqrt[3]{\frac{1}{6}}) \text{ and } -x+\pi(1+2 n)=\arcsin(\sqrt[3]{\frac{1}{6}})\]

Answer 35

now it is just two linear equations to solve

Answer 36

so just solve the last two and that will give me the soliutions?

Answer 37

well your interval is [0,2pi)
you can pick integer n such that your solutions are in the interval [0,2pi)
but yes

Answer 38

hmm those are some hard to solve equations how do i even start with
x+2npi=arcsin(3sqrt(1/6))

Answer 39

subtract 2n pi on both sides

Answer 40

I dont understand how the 2npi part can be manipulated? and huh the arcsin confuses me even more, can i just use wolframalpha or something to solve it?

Answer 41

Okay

Answer 42

now what??

Answer 43

x=arcsin(3sqrt(1/6))-2npi

Answer 44

\[\sin(x+2 n \pi)=\sqrt[3]{\frac{1}{6}} \text{ and } \sin(-x+\pi(1+2n))=\sqrt[3]{\frac{1}{6}} \\ \text{ using that \arcsin function } \\ \\ \text{ you have } \\ x+2n \pi=\arcsin(\sqrt[3]{\frac{1}{6}}) \text{ or } -x+\pi(1+2 n)=\arcsin(\sqrt[3]{\frac{1}{6}}) \\ x=-2 n \pi +\arcsin(\sqrt[3]{\frac{1}{6}}) \text{ or} -x=\arcsin(\sqrt[3]{\frac{1}{6}})-\pi(1+2n) \\ x=-2 n \pi+\arcsin(\sqrt[3]{\frac{1}{6}}) \text{ or } x=-\arcsin(\sqrt[3]{\frac{1}{6}})+\pi(1+2n)\]

Answer 45

@cutiecomittee123
Could you double check to see if the question should read:
2sin^2\(\color{red}{-}\)3sinx-1=0 or
2sin^2\(\color{red}{+}\)3sinx-1=0

Answer 46

i am going to go with @mathmate
probably a quadratic, and since both of those factor easily

Answer 47

2sin^2(x)3sin(x)-1=0 does not look like a reasonable expression for a question because it would have simplified to 6sin^3(x)-1=0

Answer 48

err lol I asked her several times @satellite73

Answer 49

i bet $7 it is one of those two

Answer 50

i double checked and they arent subtracted or added they are multiplied together

Answer 51

then i bet $10 it is a typo

Answer 52

Wait no, it is added together wow

Answer 53

@cutiecomittee123
2sin^2(x)3sin(x)-1=0 does not look like a reasonable expression for a question because it would have simplified to 6sin^3(x)-1=0
Can you post an image of the question, please?

Answer 54

Yikes lol maybe i should just reask this question

Answer 55

yes
perhaps using an alias

Answer 56

:)

Answer 57

actually what you should do is factor the expression and solve that way