How can I find all the solutions for the equation 4 sin^2(x) – 4 sin(x) + 1 = 0 in the interval from 0 to 2π?

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Finding Solutions to the Equation

To solve the equation 4 sin²(x) – 4 sin(x) + 1 = 0 within the interval [0, 2π], we can start by using a substitution method.

Step 1: Substitute

Let y = sin(x). Then, we can rewrite the equation as:

4y² – 4y + 1 = 0

Step 2: Factor or Use the Quadratic Formula

This quadratic equation can be solved using the quadratic formula, y = ( -b ± sqrt(b² – 4ac) ) / 2a, where a = 4, b = -4, and c = 1.

Now, calculate the discriminant:

D = b² – 4ac = (-4)² – 4 * 4 * 1 = 16 – 16 = 0

Since the discriminant is zero, there is only one solution for y:

y = (4 ± 0) / 8 = 0.5

Step 3: Solve for sin(x)

Now plug this value back into our substitution:

sin(x) = 0.5

Step 4: Find x in [0, 2π]

The solutions for sin(x) = 0.5 in the range [0, 2π] are:

  • x = π/6
  • x = 5π/6

Conclusion

The solutions to the equation 4 sin²(x) – 4 sin(x) + 1 = 0 in the interval from 0 to 2π are:

  • π/6
  • 5π/6

These solutions provide all the angles that satisfy the given equation within the specified interval.

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