Question

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

4 sin2x - 4 sin x + 1 = 0

so i have so far is: sin^2 x -sin x= -1/4

Answer 1

Is that, 4sin^2 x ? or is it just 4 sin (2x)?

Answer 2

it is 4 sin^2 x

Answer 3

Please edit the question so that others don't think otherwise.

Answer 4

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

What form does this look like?

Have you seen this before? \[(a-b)^2 = a^2 - 2ab + b^2\]
This is the same form, in this question though, a = 2 sin x and b = 1.

Hence, \[(2 sin x)^2 - 2(2 sin x)(1) + 1 = (2 sin x - 1)^2 = 0\]

Are you understanding this? :)

Answer 5

so sin x = sq root of 1/2 ?

Answer 6

No,\[(2 \sin x - 1)^2 = 0 ~~means~~that~~ (2 \sin x - 1) = 0 ~~right?\]

Answer 7

yes