Find all solutions in the interval [0, 2π). 4 sin2 x - 4 sin x + 1 = 0

Solution:

Using the Pythagoras theorem

x² + y² = r²

Here

cos² a + sin² a = 1

cot² a + 1 = cosec² a

1 + tan² a = sec² a

From the equation 4 sin² x - 4 sin x + 1 = 0

We know that sin x = a

So the equation is

4a² - 4a + 1 = 0

Divide the equation by 4

a² - a + 1/4 = 0

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

a = 1/2

We get

sin x = 1/2

As the interval between 0 to 2π and positive sine values are in 1st and 2nd quadrants,

x values that meet are 30° and 150° or π/6 and 5/6 π

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

Find all solutions in the interval [0, 2π). 4 sin2 x - 4 sin x + 1 = 0

Summary:

All solutions in the interval [0, 2π) for 4 sin2 x - 4 sin x + 1 = 0 are 30° and 150° or π/6 and 5/6 π.