We are given the equation:

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

### Step 1: Simplify the equation

First, divide both sides of the equation by \(\cos(x)\), assuming \(\cos(x) \neq 0\):

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

### Step 2: Solve for \(\sin(x)\)

Now, solve for \(\sin(x)\):

\[
\sin(x) = \frac{2}{4} = \frac{1}{2}
\]

### Step 3: Find the solutions

We now solve for \(x\) where \(\sin(x) = \frac{1}{2}\). The general solutions are given by:

\[
x = \arcsin\left(\frac{1}{2}\right) + 2n\pi \quad \text{or} \quad x = \pi - \arcsin\left(\frac{1}{2}\right) + 2n\pi
\]

Using \(\arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6}\), we get:

\[
x = \frac{\pi}{6} + 2n\pi \quad \text{or} \quad x = \frac{5\pi}{6} + 2n\pi
\]

where \(n\) is any integer.

### Step 4: Consider when \(\cos(x) = 0\)

Additionally, since we divided by \(\cos(x)\), we must consider when \(\cos(x) = 0\). This occurs at:

\[
x = \frac{\pi}{2} + n\pi
\]

### Final solution:

The full solution is:

\[
x = \frac{\pi}{6} + 2n\pi \quad \text{or} \quad x = \frac{5\pi}{6} + 2n\pi \quad \text{or} \quad x = \frac{\pi}{2} + n\pi
\]

where \(n\) is any integer.

Would you like further details on any steps?

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Here are 5 questions related to this:

1. How do we determine when \(\sin(x) = \frac{1}{2}\)?
2. Why do we need to consider the case when \(\cos(x) = 0\)?
3. What happens if \(\sin(x)\) equals a different value, such as \(\sin(x) = -\frac{1}{2}\)?
4. Can this method be used for other trigonometric equations?
5. How would the solution change if the equation were \(4\sin(x) \times \cos(x) = 3\cos(x)\)?

**Tip**: Always check if dividing by a term like \(\cos(x)\) might cause you to lose possible solutions. Consider the possibility that \(\cos(x) = 0\).

Discover how to solve the trigonometric equation 4sin(x) × cos(x) = 2cos(x). This step-by-step guide explains how to simplify the equation, solve for sin(x), and consider additional solutions for cos(x) = 0. Ideal for Grades 10-12 students studying trigonometry.

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