Question

Find all solutions of 4sin^2(2x)−4sin(2x)+1=0
on [0,2π).

Select all that apply.


x=5π/12
x=13π/12
x=13π/6
x=17π/6
x=π/12
x=17π/12
x=5π/6
x=π/6
There are no solutions to this equation.

Answer 1

To solve the equation \( 4\sin^2(2x) - 4\sin(2x) + 1 = 0 \), we can start by making a substitution. Let \( u = \sin(2x) \). Then, the equation becomes:

\[
4u^2 - 4u + 1 = 0
\]

This is a quadratic equation in \( u \). We can use the quadratic formula to solve for \( u \):

\[
u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 4 \cdot 1}}{2 \cdot 4}
\]

Calculating the discriminant:

\[
b^2 - 4ac = 16 - 16 = 0
\]

Since the discriminant is zero, there is exactly one solution for \( u \):

\[
u = \frac{4}{8} = \frac{1}{2}
\]

Now substituting back for \( \sin(2x) \):

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

To find \( 2x \), we need the angles where the sine is \(\frac{1}{2}\) in the interval \([0, 2\pi)\):

This occurs at:

\[
2x = \frac{\pi}{6} + 2k\pi \quad \text{and} \quad 2x = \frac{5\pi}{6} + 2k\pi
\]

for \( k \in \mathbb{Z} \).

Now we divide by 2 to solve for \( x \):

1. From \( 2x = \frac{\pi}{6} \):

\[
x = \frac{\pi}{12} + k\pi \quad (k = 0 \text{ gives } x = \frac{\pi}{12}, k = 1 \text{ gives } x = \frac{13\pi}{12})
\]

2. From \( 2x = \frac{5\pi}{6} \):

\[
x = \frac{5\pi}{12} + k\pi \quad (k = 0 \text{ gives } x = \frac{5\pi}{12}, k = 1 \text{ gives } x = \frac{17\pi}{12})
\]

Now, we check the valid solutions for \( x \) within the interval \([0, 2\pi)\):

- \( x = \frac{\pi}{12} \)
- \( x = \frac{5\pi}{12} \)
- \( x = \frac{13\pi}{12} \)
- \( x = \frac{17\pi}{12} \)

Now we double-check the potential solutions from the provided options against our results:

- \( x = \frac{\pi}{12} \) - **valid**
- \( x = \frac{5\pi}{12} \) - **valid**
- \( x = \frac{13\pi}{12} \) - **valid**
- \( x = \frac{17\pi}{12} \) - **valid**

The remaining provided options:

- \( x = \frac{13\pi}{6} \) (Invalid in \([0, 2\pi)\))
- \( x = \frac{5\pi}{6} \) (Invalid in the derived solutions)
- \( x = \frac{\pi}{6} \) (Invalid in the derived solutions)
- \( x = \frac{17\pi}{6} \) (Invalid in \([0, 2\pi)\))
- **There are no solutions to this equation.** (not valid)

### Thus, the solutions in \([0, 2\pi)\) are:

- \( x = \frac{\pi}{12} \)
- \( x = \frac{5\pi}{12} \)
- \( x = \frac{13\pi}{12} \)
- \( x = \frac{17\pi}{12} \)

In summary, the valid solutions are:

- \( x = \frac{\pi}{12} \)
- \( x = \frac{5\pi}{12} \)
- \( x = \frac{13\pi}{12} \)
- \( x = \frac{17\pi}{12} \)