Question

Find all exact solutions to 7sin2x=14sinx on [0,2π).

Answer 1

To solve the equation \( 7\sin(2x) = 14\sin(x) \), we can start by simplifying the equation.

First, divide both sides by 7 (assuming \( \sin(x) \neq 0 \)):

\[
\sin(2x) = 2\sin(x)
\]

Next, we can use the double angle identity for sine, which states that \( \sin(2x) = 2\sin(x)\cos(x) \). Substituting this into our equation gives:

\[
2\sin(x)\cos(x) = 2\sin(x)
\]

Now we can factor out \( 2\sin(x) \):

\[
2\sin(x)(\cos(x) - 1) = 0
\]

This equation gives us two factors to solve:

1. \( 2\sin(x) = 0 \)
2. \( \cos(x) - 1 = 0 \)

**Solving \( 2\sin(x) = 0 \):**

This yields:

\[
\sin(x) = 0
\]

The solutions for \( \sin(x) = 0 \) on the interval \([0, 2\pi)\) are:

\[
x = 0, \pi
\]

**Solving \( \cos(x) - 1 = 0 \):**

This yields:

\[
\cos(x) = 1
\]

The solutions for \( \cos(x) = 1 \) on the interval \([0, 2\pi)\) are:

\[
x = 0
\]

However, we have already included \( x = 0 \) in our previous solutions.

Now, we can consolidate our results. The exact solutions to the equation \( 7\sin(2x) = 14\sin(x) \) on the interval \([0, 2\pi)\) are:

\[
\boxed{0, \pi}
\]