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To solve the equation \( 2 \cos(t) = \sqrt{3} \) on the interval \([0, 2\pi]\), follow these steps:

### Step 1: Simplify the equation
First, divide both sides of the equation by 2 to isolate \( \cos(t) \):
\[
\cos(t) = \frac{\sqrt{3}}{2}
\]

### Step 2: Find the reference angle
We know from the unit circle that \( \cos(\theta) = \frac{\sqrt{3}}{2} \) occurs at the standard angles:
\[
t = \frac{\pi}{6}
\]

### Step 3: Find all solutions within \([0, 2\pi]\)
Since the cosine function is positive in the first and fourth quadrants, the solutions in the interval \([0, 2\pi]\) are:
- First quadrant: \( t = \frac{\pi}{6} \)
- Fourth quadrant: \( t = 2\pi - \frac{\pi}{6} = \frac{11\pi}{6} \)

### Final Answer:
The solutions to \( 2 \cos(t) = \sqrt{3} \) on the interval \([0, 2\pi]\) are:
\[
t = \frac{\pi}{6}, \frac{11\pi}{6}
\]

Would you like more details or have any questions about this? 

Here are some related questions:
1. How does the unit circle help in solving trigonometric equations?
2. Why are there two solutions for \( \cos(t) = \frac{\sqrt{3}}{2} \) on \([0, 2\pi]\)?
3. What would the solution be if the interval were \( [-2\pi, 2\pi] \)?
4. How can we solve \( \sin(t) = \frac{\sqrt{3}}{2} \) on \( [0, 2\pi] \)?
5. How would the solution change if the equation were \( 2 \cos(t) = -\sqrt{3} \)?

**Tip:** When solving trigonometric equations, always check the quadrant where the solution lies based on the sign of the trigonometric function.

Solve 2cos(t) = sqrt(3) on the interval [0, 2π].

Learn how to solve the trigonometric equation 2cos(t) = sqrt(3) within the interval [0, 2π]. This problem involves key trigonometric concepts, such as the unit circle and finding cosine values in the first and fourth quadrants. Suitable for Grades 10-12.

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