Find all solutions in the interval [0, 2π). 4 sin2x - 4 sin x + 1 = 0

Solution:

It is given that, equation:

4 sin²x - 4 sin x + 1 = 0,

We can say sin x = a, then the equation becomes:

4a² - 4a + 1 = 0

By dividing 4, we get,

a² - a + 1/4

So, (a - 1/2) (a - 1/2) = 0

a = 1/2

Then,

sin x = 1/2

Because the interval between 0 - 2π and positive sin values ​​are in quadrants 1 and 2 so the x values ​​that meet are 30° and 150° or π/6 and 5/6 π.

Therefore, 30° and 150° or π/6 and 5/6 π are the solutions.

Find all solutions in the interval [0, 2π). 4 sin2x - 4 sin x + 1 = 0

Summary:

All solutions in the interval [0, 2π) 4 sin²x - 4 sin x + 1 = 0 are 30° and 150° or π/6 and 5/6 π.