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- anonymous

Find all solutions in the interval [0, 2π).
sin2 x + sin x = 0

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- anonymous

Find all solutions in the interval [0, 2π).
sin2 x + sin x = 0

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- anonymous

i give medals! please help!

- anonymous

you can factor the common term of sin x
sin² x + sin x = 0
(sin x)(sin x + 1) = 0
Now set both factors equal to 0 and solve these two equations
sin x = 0
sin x + 1 = 0

- anonymous

I am very confused

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- anonymous

i only have a few more minutes to answer this please help and explain!

- Australopithecus

So you have,
\[\sin^2(x) + \sin(x) = 0\]
By the definition of a exponent
\[x^2 = x*x\]
etc
so,
\[\sin^2(x) + \sin(x) = 0\]
can be written as:
\[\sin(x)\sin(x) + \sin(x) = 0\]
Now you can factor out sin(x):
\[\sin(x)(\sin(x) + 1) = 0\]
Now look at both terms
sin(x)
and
(1 + sin(x))
When one of them equals 0 you have a solution

- anonymous

sin² x and sin x have a common term of sin x so it can be factored out giving the equation as
(sin x)(sin x + 1) = 0
If two thing multiply to be 0, one or both of them is 0. so we can write thes the two equations
sin x = 0 and sin x + 1 = 0
To solve sin x = 0, look on the unit circle and pick out all the angle where sine is 0.
To solve sin x + 1 = 0, subtract the 1 from both sides
sin x = -1
now pick out the angles that have a sine of -1.

- Australopithecus

so for
(1 + sin(x)) = 0
sin(x) must equal - 1
what value of x makes sin(x) = -1?
For the second term
sin(x) = 0
what value of x makes sin(x) equal 0
Look at the graph of sin(x)
http://www.wolframalpha.com/input/?i=graph+of+sin%28x%29+between+0+and+2pi

- Australopithecus

Remember that you are only looking for solutions for the interval [0,2pi)

- anonymous

is there answer x= 0,pi,4pi/3,5pi/3?

- anonymous

0 and pi are solutions to sin x = 0.
the other two aren't solutions to either equation. Where is sine equal to -1 between 0 and 2pi?

- anonymous

x=o,pi,pi/3,2pi/3?

- Australopithecus

when x = 0
sin(0) = 0
so x =0 is a solution
sin(4pi/3) does not equal 0 or -1 so it is not a solution
sin(5pi/3) does not equal 0 or -1 so it is not a solution
sin(pi) = 0 so it is a solution

- Australopithecus

Look at the interval 0 to 2pi
It is just the top section of the unit circle

- anonymous

0, pi, 3pi/2

- Australopithecus

yup

- anonymous

Thanks!

- Australopithecus

If you found my help useful please rate me using the qualified help located at the top

- Australopithecus

So I can get paid :)

- anonymous

I will!

- Australopithecus

Not that I wouldnt have helped you for free :)

- anonymous

How do you do that?

- Australopithecus

http://i.imgur.com/7yfPY5A.jpg

- Australopithecus

Click this button :)

- Australopithecus

Ha brain is in stupid mode right now Click rate qualified helper so the button next to it

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