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To solve the equation
\[
(2\sin x - 1)(\cos x + 3) = 0
\]
on the interval \([0, 2\pi]\), we can use the fact that if a product of two factors is zero, then at least one of the factors must be zero. Therefore, we split the equation into two cases:

1. \(2\sin x - 1 = 0\)
2. \(\cos x + 3 = 0\)

### Step 1: Solve \(2\sin x - 1 = 0\)
This equation simplifies to:
\[
\sin x = \frac{1}{2}
\]

In the interval \([0, 2\pi]\), \(\sin x = \frac{1}{2}\) at the following values of \(x\):
\[
x = \frac{\pi}{6}, \quad x = \frac{5\pi}{6}
\]

### Step 2: Solve \(\cos x + 3 = 0\)
This equation simplifies to:
\[
\cos x = -3
\]

However, since the range of \(\cos x\) is \([-1, 1]\), \(\cos x = -3\) has no solutions. Therefore, there are no values of \(x\) that satisfy this part of the equation.

### Final Solution
The solutions to the equation \((2\sin x - 1)(\cos x + 3) = 0\) on the interval \([0, 2\pi]\) are:
\[
x = \frac{\pi}{6}, \frac{5\pi}{6}
\]

Would you like a deeper explanation or have any questions?

Here are five related questions to extend your understanding:

1. How would the solutions change if the interval were \([-2\pi, 2\pi]\)?
2. What if the equation were \((2\sin x - 1)(\cos x - 1) = 0\)?
3. How can we solve similar trigonometric equations with \(\tan x\)?
4. Why does the range of \(\cos x\) affect possible solutions for \(\cos x = -3\)?
5. How can we verify the solutions using a unit circle?

**Tip:** For trigonometric equations, it’s useful to keep the unit circle and basic angles in mind to quickly identify possible solutions.

Find all solutions of the equation (2sinx - 1)(cosx + 3) = 0 in the interval [0, 2π].

Learn how to solve the trigonometric equation (2sinx - 1)(cosx + 3) = 0 in the interval [0, 2π]. This step-by-step guide explains how to use the Zero Product Property and solve for sine and cosine values. Suitable for students in Grades 10-12.

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