Question

Find all solutions in the interval [0, 2π).

4 sin2x - 4 sin x + 1 = 0

Answer 1

@Luigi0210

Answer 2

Do you think that's factorable?

Answer 3

no.. becasue 1x4=4 and -1x-4=4 but -1-4=-5 and 1+4=5 so definitely no

Answer 4

Hmm, try the quadratic formula?

Answer 5

+4sinx +-sqrt-4sinx^2-4(4sin2x)(1) all over 2(4sin2x)

Answer 6

Nope, just the constant, not the whole thing.

Answer 7

what?

Answer 8

like 4+-sqrt-4^2-4(4)(1) all over 2(4)?

Answer 9

Yes, and also you can make equations with http://prntscr.com/38jzw9

Answer 10

Makes it neater :)

Answer 11

\[( 4+-\sqrt{0})/8 \]

Answer 12

Right! So that gives us \(x=.5\). Can you work backwards to get the original equation?

Answer 13

(ノ_・。)
.5 isnt an answer.....

Answer 14

Because we're not done yet silly >.>

Answer 15

wait then what am i working backwards from? and i thought the sqrt of 0 was impossible

Answer 16

To get the equations \((2sinx-1)(2sinx-1)=0\)

Answer 17

where did you get that out of (4+-sqrt0)/8

Answer 18

When you simplify \( \frac{4 \pm \sqrt{0}}{8}\) you get \(\frac{1}{2}\) when you simplify, right?

Answer 19

i just went up a level for classifying a triangle (ﾉ◕ヮ◕)ﾉ*:･ﾟ✧

Answer 20

no, 4/8=1/2 and sqrt0/8= i have no idea

Answer 21

Congrats :)
And you're thinking is rational. \(\sqrt{0}=0\) so \(\frac{\sqrt{0}}{8}=0\)
So all you have is 1/2

Answer 22

But you can't separate them, you have to simplify the square root first.

Answer 23

Sorry, confusing you again? .-.

Answer 24

yeah you can... as long as theres a plus or minus inbetween...

Answer 25

Sorry, used to doing it all as one.. but yea. \(\sqrt{0}=0\)
And \(4 \pm 0\) will give you \(4\)

Answer 26

so i was right (⊙ヮ⊙)

Answer 27

Yea you were xD

I'm trying to think of what to do next..

Answer 28

Wait, nvm I know, I was thinking too hard!
So we have the equation \((2sinx-1)^2=0\) Can you solve that?

Answer 29

exactly! im about to chuck my computer at a wall.... (2sinx-1)(2sinx-1)= 4sinx^2-4sinx+1

Answer 30

Is your computer hatin on you? D:

And no, SOLVE, not DISTRIBUTE >.<

Answer 31

So take the square root of both sides:
\(2sinx-1=0\), now add one and divide. \(sinx=\frac{1}{2}\)
Now, just find the values of sin that make it equal 1/2 and you're done :)

Answer 32

i did /).(\ i used foil

Answer 33

Don't use FOIL, that'll undo all our work.. we have to set it equal to zero and solve for x.

Answer 34

Using your interval given.. \(\LARGE x=\frac{\pi}{3}\) and \(\LARGE x=\frac{2\pi}{3}\)