Solving $\sqrt{3}\sin(x) + \cos(x) - 2 = 0$: A Step-by-Step Guide

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Hey everyone! Let's dive into solving the trigonometric equation $\sqrt{3}\sin(x) + \cos(x) - 2 = 0$. This might seem a bit tricky at first, but trust me, we can break it down step-by-step. We'll go through a few different approaches, including the one you started with, and see how they pan out. This is gonna be fun, so buckle up!

Understanding the Problem and Initial Approaches

So, our main goal here is to find the values of x that make the equation $\sqrt{3}\sin(x) + \cos(x) - 2 = 0$ true. This is a classic trigonometry problem, and there are a few different ways we can approach it. One of the most common strategies is to try and simplify the equation, maybe by using some trigonometric identities or algebraic manipulations. You mentioned that you tried separating the radical and squaring, which is a valid idea, but sometimes it can introduce extraneous solutions, so we need to be careful.

First, let's rearrange the equation to make it easier to work with. We have:

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

One initial thought might be to try and express the left-hand side as a single trigonometric function. We can do this by using the angle sum identity. Recall the identity:

$\sin(A + B) = \sin(A)\cos(B) + \cos(A)\sin(B)$

We want to rewrite $\sqrt{3}\sin(x) + \cos(x)$ in a similar form. Let's try to find an angle $\alpha$ such that $\cos(\alpha) = \frac{\sqrt{3}}{2}$ and $\sin(\alpha) = \frac{1}{2}$. This would allow us to rewrite the left side in the form of $\sin(x + \alpha)$ or $\cos(x + \alpha)$. That angle is $\alpha = \frac{\pi}{6}$ (or 30 degrees).

So, let's rewrite our equation using this angle. Multiplying and dividing the left side by 2, we get:

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

This can be rewritten as:

$2(\cos(\frac{\pi}{6})\sin(x) + \sin(\frac{\pi}{6})\cos(x)) = 2$

Using the angle sum identity for sine, we have:

$2\sin(x + \frac{\pi}{6}) = 2$

Now, we can easily solve for x.

Solving Using Trigonometric Identities and the Angle Sum Identity

Let's continue with the approach we started, focusing on the angle sum identity. As we derived in the previous section, we have:

$2\sin(x + \frac{\pi}{6}) = 2$

Now, divide both sides by 2:

$\sin(x + \frac{\pi}{6}) = 1$

Remember, the sine function equals 1 at $\frac{\pi}{2}$ plus any multiple of $2\pi$. Therefore:

$x + \frac{\pi}{6} = \frac{\pi}{2} + 2n\pi$, where n is an integer.

To solve for x, subtract $\frac{\pi}{6}$ from both sides:

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

$x = \frac{3\pi}{6} - \frac{\pi}{6} + 2n\pi$

$x = \frac{2\pi}{6} + 2n\pi$

$x = \frac{\pi}{3} + 2n\pi$

So, the general solution for x is $\frac{\pi}{3} + 2n\pi$. This means that x can be $\frac{\pi}{3}$, $\frac{\pi}{3} + 2\pi$, $\frac{\pi}{3} - 2\pi$, and so on. These are all the angles where the original equation holds true. This is the most direct and elegant solution!

Key takeaway: By recognizing the structure of the equation and using trigonometric identities strategically, we can simplify the problem and arrive at the solution efficiently. Using the angle sum identity is often a great way to combine sine and cosine terms.

Your Attempt and the Quadratic Equation Method

You mentioned that you separated the radical (which isn't quite accurate here, but I understand what you mean) and then squared the equation and noted $\cos(x) = t$, resulting in a quadratic equation. Let's explore that path and see what happens, though we already have a straightforward solution. This method is a bit more involved, and as I mentioned before, squaring equations can introduce extraneous solutions. It's important to always check your solutions when using this method.

Starting with our original equation:

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

Let's isolate the cosine term:

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

Now, square both sides:

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

Expand the right side:

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

Recall that $\cos^2(x) = 1 - \sin^2(x)$. Substitute this into the equation:

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

Rearrange the terms to form a quadratic equation in terms of $\sin(x)$:

$0 = 4\sin^2(x) - 4\sqrt{3}\sin(x) + 3$

Now, let $t = \sin(x)$. Our equation becomes:

$4t^2 - 4\sqrt{3}t + 3 = 0$

This is a quadratic equation that we can solve using the quadratic formula:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case, $a = 4$, $b = -4\sqrt{3}$, and $c = 3$. Plugging in these values, we get:

$t = \frac{4\sqrt{3} \pm \sqrt{(-4\sqrt{3})^2 - 4(4)(3)}}{2(4)}$

$t = \frac{4\sqrt{3} \pm \sqrt{48 - 48}}{8}$

$t = \frac{4\sqrt{3}}{8}$

$t = \frac{\sqrt{3}}{2}$

Since $t = \sin(x)$, we have $\sin(x) = \frac{\sqrt{3}}{2}$. The solutions to this equation are $x = \frac{\pi}{3} + 2n\pi$ and $x = \frac{2\pi}{3} + 2n\pi$. Now the crucial step: we must check these solutions in the original equation because we squared the equation, which can introduce extraneous solutions.

For $x = \frac{\pi}{3}$: $\sqrt{3}\sin(\frac{\pi}{3}) + \cos(\frac{\pi}{3}) = \sqrt{3}(\frac{\sqrt{3}}{2}) + \frac{1}{2} = \frac{3}{2} + \frac{1}{2} = 2$. This solution works!

For $x = \frac{2\pi}{3}$: $\sqrt{3}\sin(\frac{2\pi}{3}) + \cos(\frac{2\pi}{3}) = \sqrt{3}(\frac{\sqrt{3}}{2}) - \frac{1}{2} = \frac{3}{2} - \frac{1}{2} = 1$. This solution does not work!

So, the only valid solution from this method is $x = \frac{\pi}{3} + 2n\pi$, which is the same as what we found before. This confirms that the method works but emphasizes the importance of checking the results.

Why the Cauchy-Schwarz Inequality Doesn't Directly Apply (In This Case)

You also mentioned the Cauchy-Schwarz Inequality. While the Cauchy-Schwarz Inequality is a powerful tool, it's not directly applicable in a straightforward way to solve this equation. However, it can be used to understand the range of values that the expression $\sqrt{3}\sin(x) + \cos(x)$ can take. Let's see how it works.

The Cauchy-Schwarz Inequality states that for any real numbers $a_1, a_2, b_1, b_2$:

$(a_1^2 + a_2^2)(b_1^2 + b_2^2) \ge (a_1b_1 + a_2b_2)^2$

Let $a_1 = \sqrt{3}$, $a_2 = 1$, $b_1 = \sin(x)$, and $b_2 = \cos(x)$. Applying the Cauchy-Schwarz Inequality, we get:

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

Since $\sin^2(x) + \cos^2(x) = 1$, this simplifies to:

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

$4 \ge (\sqrt{3}\sin(x) + \cos(x))^2$

Taking the square root of both sides, we get:

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

This tells us that the expression $\sqrt{3}\sin(x) + \cos(x)$ can only take values between -2 and 2. In our equation, we set it equal to 2. Thus we can conclude that if a solution exists, it can only have values within the boundary of the Cauchy Schwarz result. It's a nice way to check if our answer makes sense, ensuring it is within the bounds and doesn't exceed the maximum value possible. This can be useful for verifying the solutions we obtain using other methods. The equality case in the Cauchy-Schwarz Inequality holds when $\frac{a_1}{b_1} = \frac{a_2}{b_2}$, meaning in our case, when $\frac{\sqrt{3}}{\sin(x)} = \frac{1}{\cos(x)}$. This occurs when the angle is in the first quadrant with a $y$ coordinate of $\frac{\sqrt{3}}{2}$ and an $x$ coordinate of $\frac{1}{2}$, as we found earlier.

Key takeaway: Cauchy-Schwarz helps us understand the range of the function but isn't a direct method for solving the equation itself. It provides a way to check our solutions.

Conclusion: Putting It All Together

Alright, we've explored a few different paths to solve $\sqrt{3}\sin(x) + \cos(x) - 2 = 0$. The most efficient and elegant method involves using the angle sum identity to rewrite the left side and then solving for x. We also examined your approach, using the quadratic equation, which, while valid, required careful checking of the solutions to avoid extraneous ones. The Cauchy-Schwarz inequality gives us valuable context by helping us understand the range of the function. The critical idea to remember is the strategic use of trigonometric identities to simplify complex expressions and bring them to a more manageable form.

So, in the end, the general solution to this equation is $x = \frac{\pi}{3} + 2n\pi$. I hope this step-by-step guide was helpful! Keep practicing, and you'll become a trigonometry pro in no time! If you have any other questions, feel free to ask. Happy solving, everyone!