We are tasked with solving the trigonometric equation:

$\sin(x) + \sqrt{3} \cos(x) = 1$

where $x$ is an element of the interval $[0, 2\pi]$.

Step 1: Rewriting the Equation

We can combine the sine and cosine terms into a single trigonometric function. To do this, let's express the equation in the form $R \sin(x + \alpha)$. First, recall the identity:

$R \sin(x + \alpha) = R (\sin(x)\cos(\alpha) + \cos(x)\sin(\alpha))$

To match the original equation $\sin(x) + \sqrt{3} \cos(x)$, we want:

$R \cos(\alpha) = 1 \quad \text{and} \quad R \sin(\alpha) = \sqrt{3}$

Step 2: Solving for $R$ and $\alpha$

From these two relationships, we can find $R$ and $\alpha$.

$R = \sqrt{(1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2$

Now, to find $\alpha$:

$\tan(\alpha) = \frac{\sqrt{3}}{1} = \sqrt{3}$

This implies:

$\alpha = \frac{\pi}{3}$

Step 3: Substituting Back

We can now rewrite the original equation as:

$2 \sin(x + \frac{\pi}{3}) = 1$

Dividing both sides by 2:

$\sin(x + \frac{\pi}{3}) = \frac{1}{2}$

Step 4: Solving for $x$

The general solution to $\sin(\theta) = \frac{1}{2}$ is:

$\theta = \frac{\pi}{6} + 2k\pi \quad \text{or} \quad \theta = \pi - \frac{\pi}{6} + 2k\pi$

Substitute $x + \frac{\pi}{3}$ for $\theta$:

$x + \frac{\pi}{3} = \frac{\pi}{6} + 2k\pi \quad \text{or} \quad x + \frac{\pi}{3} = \frac{5\pi}{6} + 2k\pi$

Now, solve for $x$:

1. $x = \frac{\pi}{6} - \frac{\pi}{3} + 2k\pi = -\frac{\pi}{6} + 2k\pi$
2. $x = \frac{5\pi}{6} - \frac{\pi}{3} + 2k\pi = \frac{\pi}{2} + 2k\pi$

Step 5: Validating within the interval $[0, 2\pi]$

We need the values of $x$ that fall within the interval $[0, 2\pi]$.

For $x = -\frac{\pi}{6} + 2k\pi$, take $k = 1$:

$x = -\frac{\pi}{6} + 2\pi = \frac{11\pi}{6}$

For $x = \frac{\pi}{2} + 2k\pi$, take $k = 0$:

$x = \frac{\pi}{2}$

Final Solution:

The values of $x$ in the interval $[0, 2\pi]$ are:

$x = \frac{\pi}{2}, \quad x = \frac{11\pi}{6}$

Would you like more details or have any questions?

Here are 5 related questions:

1. How do you convert trigonometric equations into simpler forms?
2. What is the general solution for $\sin(\theta) = \text{constant}$?
3. How can you apply phase shifts in trigonometric identities?
4. How can the Pythagorean theorem help find $R$ in these problems?
5. How do you validate trigonometric solutions within a specific interval?

Tip: Remember to always check that your solutions fall within the given interval when solving trigonometric equations.