Question

Can someone please help me with this problem. I have no idea where to start.

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

4 sin2 x - 4 sin x + 1 = 0

pi divided by three, five pi divided by three

pi divided by six, eleven pi divided by six

seven pi divided by six, eleven pi divided by six

pi divided by six, five pi divided by six

Answer 1

PLEASE HELP

Answer 2

\[4\sin2x-4\sin x+1=0\]
or
\[4\sin^2x-4\sin x+1=0~~?\]

Answer 3

@SithsAndGiggles did u ever get the answer to this?

Answer 4

Technically no, because my question was never addressed. If it's the second equation I wrote, it's pretty simple.

Answer 5

yes it's the second one :)

Answer 6

Well we have an equation that's quadratic in \(\sin x\). What that means is that if we replace \(\sin x\) with \(t\), we get
\[4t^2-4t+1=0\]
which can be solved for \(t\) using the quadratic formula.

So say we find our solution(s), then we would have to solve \(\sin x=\text{solutions for t}\) for \(x\).

Answer 7

okay how would we do that?

Answer 8

\[4t^2-4t+1=0~~\implies~~t=\frac{4\pm\sqrt{16-16}}{8}=\frac{1}{2}\]
Replace \(t\) with \(\sin x\), then you have
\[\sin x=\frac{1}{2}~~\implies~~x=\arcsin\frac{1}{2}\]
In the interval \([0,2\pi)\), you have two corresponding solutions, \(x=\dfrac{\pi}{6}\) and \(x=\dfrac{5\pi}{6}\).

Answer 9

wow! thank you so much that was great! i get it now I really appreciate it!

Answer 10

yw