Question

Please help! solve: 4sin³x+2sin²x-2sinx-1=0 for 0° ≤ x ≤ 360°

Answer 1

can you take common factor on the left-hand side?

Answer 2

common factor would be sin(x) right?

Answer 3

... h... actually... after getting rid of the 1, yes

Answer 4

well... 2sin(x)

Answer 5

You add 1 to the right side but what would the left hand side look like after you simplify the gcf

Answer 6

\(\bf 4sin^3(x)+2sin^2(x)-2sin(x)-1=0\\ \quad \\

4sin^3(x)+2sin^2(x)-2sin(x)=1\\ \quad \\
2sin(x)\quad [2sin^2(x)+sin(x)-1] = 1\)

Answer 7

hmmmm

Answer 8

So are you looking to solve 4u^3+2u^2-2u-1=0

Did you try to find the possible 0's?

Answer 9

yea... I think so @myininaya is correct.. you may have to factor it with the 0 on the right-hand side

Answer 10

or you could factor by grouping :) much simpler

Answer 11

Yes I am trying to solve for the possible solutions in degrees. How would you factor by grouping?

Answer 12

\[(4\sin^3(x)+2\sin^2(x))+(-2\sin(x)-1)=0\]

Answer 13

Look at those first two terms. They have common factor.
Look at the last terms. Factor out -1.

Answer 14

Or you can look at it in terms of u instead of sin at first. Your choice.

Answer 15

would it be 2sin^3(x) +sin(x)) + (2sin(x))=0 ?

Answer 16

\[(4u^3+2u^2)+(-2u-1)=0\]

Answer 17

What does 4u^3 and 2u^2 have in common?

Answer 18

2u^2? or just u^2?

Answer 19

The gcf would be 2u^2
So we have
\[2u^2(2u+1)+(-2u-1)=0\]
Those last two terms, you can factor out -1.

Answer 20

2u^2(2u+1) + ( 2sin) = 0

Answer 21

(-2u-1)=-(2u+1)

Answer 22

So (-2sin(x)-1)=-(2sin(x)+1)

Answer 23

\[2u^2(2u+1)-1(2u+1)=0\]

Now you have two terms.
(2u+1) is a factor of both of those terms.
Do you think you can completely factor the left hand side now.

Answer 24

2u^2-1=0

Answer 25

Don't forget about the (2u+1)

So to factor \[2u^2(2u+1)-1(2u+1) \text{ we find what both of our terms have in common } \]
The (2u+1)
So we factor the (2u+1) out giving us
\[(2u+1)(2u^2-1)=0\]
Set both factors equal to 0.
Don't forget u is sin(x).

Answer 26

Oh yeah that's right, I haven't done factoring in awhile. what would be the next step after this one?

Answer 27

Setting both factors equal to 0.

Answer 28

sin x +1 = 0
sin x = -1
and
sin^2x-1=0
sin^2x = 1

Answer 29

Your 2's are disappearing?

Answer 30

\[2u+1=0 \text{ or } 2u^2-1=0\]
\[2u=-1 \text{ or } 2u^2=1\]
\[u=\frac{-1}{2} \text{ or } u^2=\frac{1}{2}\]

Answer 31

How would you convert that to degrees for the final answer?

Answer 32

sin(x)=-1/2
Did I just show you the unit circle?

Answer 33

is it 210 and 330 degrees?

Answer 34

You also need to solve sin^2(x)=1/2
Take the square root of both sides
\[\sin(x)=\pm \sqrt{ \frac{1}{2} }\]
Don't forget to solve these two as well.
And yes those are the solutions to sin(x)=-1/2

Answer 35

Now for this one, it maybe easy for you to rationalize the denominator first before looking on the unit circle.

Answer 36

so the four answers would be 30, 150, 210, and 330 degrees?

Answer 37

\[\sin(x)=\frac{\sqrt{2}}{2} \text{ or } \sin(x)=-\frac{\sqrt{2}}{2}\]
I don't think you solved these correctly.

Answer 38

You solved sin(x)=-1/2 correctly.
So part of your answer is 210 and 330 degrees.

Answer 39

okay thanks for the help

Answer 40

Np. You got it from here?

Answer 41

yeah i got it.

Answer 42

Neatness.