The equation given is:

$\cos(2x) - \sqrt{3} \sin(x) = 1$

We are tasked with finding all solutions for $x$ in the interval $[0, 2\pi]$.

Step-by-step solution:

1. Rearrange the equation:

   $\cos(2x) = 1 + \sqrt{3} \sin(x)$

2. Substitute $\cos(2x)$ using a double-angle identity: Recall that $\cos(2x)$ can be expressed as:

   $\cos(2x) = 1 - 2\sin^2(x)$

   Substituting this into the equation:

   $1 - 2\sin^2(x) = 1 + \sqrt{3} \sin(x)$

3. Simplify the equation: First, subtract 1 from both sides:

   $-2\sin^2(x) = \sqrt{3}\sin(x)$

   Divide by $-1$ to simplify further:

   $2\sin^2(x) = -\sqrt{3} \sin(x)$

4. Factor the equation: Move all terms to one side to set the equation to zero:

   $2\sin^2(x) + \sqrt{3}\sin(x) = 0$

   Factor out $\sin(x)$:

   $\sin(x)(2\sin(x) + \sqrt{3}) = 0$

5. Solve each factor: Now, we have two cases:

   • Case 1: $\sin(x) = 0$

     The solutions are $x = 0, \pi, 2\pi$.

   • Case 2: $2\sin(x) + \sqrt{3} = 0$

     Solve for $\sin(x)$:

     $2\sin(x) = -\sqrt{3}$ $\sin(x) = -\frac{\sqrt{3}}{2}$

     The solutions for $\sin(x) = -\frac{\sqrt{3}}{2}$ in the interval $[0, 2\pi]$ are:

     $x = \frac{4\pi}{3}, \frac{5\pi}{3}$

Final answer:

The solutions in the interval $[0, 2\pi]$ are:

$x = 0, \pi, 2\pi, \frac{4\pi}{3}, \frac{5\pi}{3}$

Would you like further details or explanations for any of the steps?

Here are 5 related questions to consider:

1. What are the common trigonometric identities used to simplify trigonometric equations?
2. How does the double-angle identity for cosine help in solving equations like this?
3. Can you explain how factoring helps in solving quadratic trigonometric equations?
4. What is the significance of the interval $[0, 2\pi]$ when finding solutions?
5. How do you determine solutions for sine or cosine in different quadrants?

Tip: Always look for opportunities to apply trigonometric identities when working with equations involving multiple angles (like $2x$ or $3x$).