We are tasked with solving the equation:

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

for $x$ in the interval $[0, 2\pi]$. We can solve this using the "R formula" (also known as the method of expressing a linear combination of sine and cosine as a single trigonometric function).

Step 1: Express the equation in the form $R \cos(x - \alpha) = \sqrt{2}$

We begin by writing the left-hand side $\cos(x) - \sqrt{3} \sin(x)$ as a single cosine term. The general method is:

$R \cos(x - \alpha) = R \left( \cos(x) \cos(\alpha) + \sin(x) \sin(\alpha) \right)$

We want to match the coefficients of $\cos(x)$ and $\sin(x)$ with those in the equation $\cos(x) - \sqrt{3} \sin(x)$.

Thus, we have:

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

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

1. Find $R$:

From the Pythagorean identity, we can find $R$ as:

$R = \sqrt{ \cos^2(\alpha) + \sin^2(\alpha) } = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1 + 3} = \sqrt{4} = 2$

2. Find $\alpha$:

Next, we calculate $\alpha$ using the following relations:

$\cos(\alpha) = \frac{1}{R} = \frac{1}{2}, \quad \sin(\alpha) = \frac{-\sqrt{3}}{R} = \frac{-\sqrt{3}}{2}$

From these, we recognize that:

$\alpha = \frac{5\pi}{6} \quad (\text{since the cosine is positive and the sine is negative, indicating the second quadrant}).$

Step 3: Rewrite the equation

Substituting $R = 2$ and $\alpha = \frac{5\pi}{6}$ into the equation, we now have:

$2 \cos\left(x - \frac{5\pi}{6}\right) = \sqrt{2}$

Step 4: Solve for $x$

Now, divide both sides by 2:

$\cos\left(x - \frac{5\pi}{6}\right) = \frac{\sqrt{2}}{2}$

We know that $\cos(\theta) = \frac{\sqrt{2}}{2}$ has solutions:

$\theta = \pm \frac{\pi}{4} + 2k\pi \quad \text{for integer } k$

Thus, for our equation:

$x - \frac{5\pi}{6} = \pm \frac{\pi}{4} + 2k\pi$

Step 5: Solve for $x$

1. For $x - \frac{5\pi}{6} = \frac{\pi}{4} + 2k\pi$:

$x = \frac{\pi}{4} + \frac{5\pi}{6} + 2k\pi = \frac{3\pi}{12} + \frac{10\pi}{12} + 2k\pi = \frac{13\pi}{12} + 2k\pi$

2. For $x - \frac{5\pi}{6} = -\frac{\pi}{4} + 2k\pi$:

$x = -\frac{\pi}{4} + \frac{5\pi}{6} + 2k\pi = -\frac{3\pi}{12} + \frac{10\pi}{12} + 2k\pi = \frac{7\pi}{12} + 2k\pi$

Step 6: Find the solutions in the interval $[0, 2\pi]$

Now, we check the solutions for $k = 0$ and $k = 1$:

• For $x = \frac{13\pi}{12}$, this is between 0 and $2\pi$, so it's valid.
• For $x = \frac{7\pi}{12}$, this is also between 0 and $2\pi$, so it's valid.
• For $x = \frac{13\pi}{12} + 2\pi = \frac{37\pi}{12}$, this is greater than $2\pi$, so it's outside the range.

Thus, the valid solutions in the range $[0, 2\pi]$ are:

$x = \frac{13\pi}{12}, \quad x = \frac{7\pi}{12}$

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Would you like further clarification on any of the steps? Here are some related questions for you:

1. How do you convert between different forms of trigonometric equations?
2. What are other methods for solving trigonometric equations besides using the R formula?
3. How can you determine the general solution for periodic trigonometric functions?
4. What role do the quadrant signs play in determining solutions to trigonometric equations?
5. How do you use the unit circle to solve trigonometric equations like this?

Tip: Always check the solutions are within the specified range, especially when dealing with periodic functions like sine and cosine!