sin(x)+sqrt(3)*cos(x)=1 der x er et element i fra 0 til 2 pi

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.

sin(x) + sqrt(3)*cos(x) = 1 der x er et element i fra 0 til 2 pi

Step-by-step solution to the trigonometric equation sin(x) + sqrt(3)*cos(x) = 1, where x is in the interval [0, 2π]. This detailed explanation covers key trigonometric concepts like phase shifts, the sum of angles identity, and solving for sine with phase transformations. Suitable for students in Grades 10-12.

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