How can we find all solutions to the equation 4sin²(x) – 4sin(x) + 1 = 0 in the interval [0, 2π]?

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To solve the equation 4sin²(x) – 4sin(x) + 1 = 0 in the interval [0, 2π], we will first treat it as a quadratic equation in terms of sin(x).

Let us substitute y = sin(x). The equation then becomes:

4y² – 4y + 1 = 0

This can be solved using the quadratic formula:

y = \frac{-b \pm \sqrt{b² – 4ac}}{2a}

Here, a = 4, b = -4, and c = 1. Plugging in these values:

y = \frac{-(-4) \pm \sqrt{(-4)² – 4 \cdot 4 \cdot 1}}{2 \cdot 4}

Calculating the discriminant:

(-4)² – 4 \cdot 4 \cdot 1 = 16 – 16 = 0

Since the discriminant is 0, we have a single (repeated) root:

y = \frac{4}{8} = \frac{1}{2}

Now, substituting back for sin(x), we get:

sin(x) = \frac{1}{2}

To find all solutions for sin(x) = \frac{1}{2}, we look for angles in the interval [0, 2π]. The angle for which sine takes the value \frac{1}{2} is:

  • x = \frac{π}{6} (in the first quadrant)
  • x = \frac{5π}{6} (in the second quadrant)

Thus, the solutions to the equation 4sin²(x) – 4sin(x) + 1 = 0 in the interval [0, 2π] are:

  • x = \frac{π}{6}
  • x = \frac{5π}{6}

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