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

**Solution:**

Using the Pythagoras theorem

x^{2} + y^{2} = r^{2}

Here

cos^{2} a + sin^{2} a = 1

cot^{2} a + 1 = cosec^{2} a

1 + tan^{2} a = sec^{2} a

From the equation 4 sin^{2} x - 4 sin x + 1 = 0

We know that sin x = a

So the equation is

4a^{2} - 4a + 1 = 0

Divide the equation by 4

a^{2} - 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.

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