Solving Trigonometric Equations: A Step-by-Step Guide

Hey guys, let's dive into solving the trigonometric equation: ${\sqrt{3}\sin(x) + \cos(x) - 2 = 0}$. This equation might seem a bit intimidating at first, but trust me, we can break it down into manageable steps. We'll explore different approaches to find the solutions for x. One way involves a little bit of algebraic manipulation and using trigonometric identities. Another interesting method uses the Cauchy-Schwarz Inequality, offering a unique perspective on solving this kind of problem. Let's get started!

Initial Approaches and Understanding the Problem

So, the first thing we want to do when we come across an equation like this is to try and get a feel for it. Remember, the goal is to find the values of x that satisfy the equation. In this case, we have a mix of sine and cosine functions, which can make things a bit tricky. A good starting point is often to try and isolate one of the trigonometric functions and then square both sides. However, we must be careful because squaring can introduce extraneous solutions, meaning solutions that don't actually work in the original equation. We'll need to check our answers at the end.

Let's start by rearranging the equation to isolate the radical term. We have:

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

Now, square both sides:

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

Expanding the right side gives:

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

Now we can use the Pythagorean identity ${\sin^2(x) + \cos^2(x) = 1}$ to replace ${\sin^2(x)}$ with ${1 - \cos^2(x)}$:

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

Simplifying this gives us a quadratic equation in terms of ${\cos(x)}$:

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

${4\cos^2(x) - 4\cos(x) + 1 = 0}$

This is great! We've transformed our original equation into a much more manageable quadratic equation. Let's solve it and see where it leads us.

Solving the Quadratic Equation

Great, now we have a quadratic equation! Let's solve it. We can make our lives a bit easier by substituting ${t = \cos(x)}$. Then our equation becomes:

${4t^2 - 4t + 1 = 0}$

This quadratic equation is pretty easy to solve. You can either factor it or use the quadratic formula. In this case, factoring is straightforward:

${(2t - 1)^2 = 0}$

This means that ${2t - 1 = 0}$, which gives us:

${t = \frac{1}{2}}$

Since ${t = \cos(x)}$, we now have:

${\cos(x) = \frac{1}{2}}$

Now, we need to find the values of x that satisfy this equation. Thinking about the unit circle, the cosine function is equal to 1/2 at two main angles within the range of ${0}$ to ${2\pi}$: ${\frac{\pi}{3}}$ and ${\frac{5\pi}{3}}$. But hold on a sec, we need to consider if there are any extraneous solutions introduced by squaring the original equation. Always, always check!

Checking for Extraneous Solutions and Finding the Correct Solutions

As we noted earlier, squaring both sides of an equation can sometimes lead to extraneous solutions. So, before we celebrate, we need to check our solutions to make sure they work in the original equation: ${\sqrt{3}\sin(x) + \cos(x) - 2 = 0}$.

Let's start with ${x = \frac{\pi}{3}}$:

${\sqrt{3}\sin(\frac{\pi}{3}) + \cos(\frac{\pi}{3}) - 2 = \sqrt{3} \cdot \frac{\sqrt{3}}{2} + \frac{1}{2} - 2 = \frac{3}{2} + \frac{1}{2} - 2 = 0}$

This solution works! Great!

Now let's check ${x = \frac{5\pi}{3}}$:

${\sqrt{3}\sin(\frac{5\pi}{3}) + \cos(\frac{5\pi}{3}) - 2 = \sqrt{3} \cdot (-\frac{\sqrt{3}}{2}) + \frac{1}{2} - 2 = -\frac{3}{2} + \frac{1}{2} - 2 = -3}$

Oops! This solution does not work. It is an extraneous solution that we introduced by squaring the equation earlier. Therefore, ${x = \frac{5\pi}{3}}$ is not a solution.

So, the only solution in the interval ${[0, 2\pi)}$ is ${x = \frac{\pi}{3}}$. Generally, the complete solution set is:

${x = \frac{\pi}{3} + 2n\pi}$, where n is an integer.

Using Cauchy-Schwarz Inequality

Alright, let's explore another cool way to solve this problem. This method uses the Cauchy-Schwarz Inequality. If you're not familiar with it, don't worry, we will break it down. The Cauchy-Schwarz Inequality is a powerful tool that has many applications in mathematics, and it can be used to solve trigonometric equations too! Specifically, it states that for any real numbers ${a_1, a_2, ..., a_n}$ and ${b_1, b_2, ..., b_n}$:

${(a_1^2 + a_2^2 + ... + a_n^2)(b_1^2 + b_2^2 + ... + b_n^2) \ge (a_1b_1 + a_2b_2 + ... + a_nb_n)^2}$

Let's rewrite our equation ${\sqrt{3}\sin(x) + \cos(x) = 2}$. Now, let's think of ${\sqrt{3}\sin(x) + \cos(x)}$ as the dot product of two vectors. Specifically, consider the vectors ${\langle \sqrt{3}, 1 \rangle}$ and ${\langle \sin(x), \cos(x) \rangle}$. Then, using the dot product formula:

${\langle \sqrt{3}, 1 \rangle \cdot \langle \sin(x), \cos(x) \rangle = \sqrt{3}\sin(x) + \cos(x)}$

Now we can apply the Cauchy-Schwarz Inequality. Let ${a = \langle \sqrt{3}, 1 \rangle}$ and ${b = \langle \sin(x), \cos(x) \rangle}$. Then:

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

Simplifying, we get:

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

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

Taking the square root of both sides:

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

Which means:

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

So, if ${\sqrt{3}\sin(x) + \cos(x) = 2}$, it means that the equality condition in the Cauchy-Schwarz Inequality must hold. Equality holds if and only if the vectors ${a}$ and ${b}$ are proportional. This means there exists a constant ${k}$ such that ${\sin(x) = k\sqrt{3}}$ and ${\cos(x) = k}$.

Applying Cauchy-Schwarz Inequality to Solve

Since we've established that equality in the Cauchy-Schwarz inequality holds, we know that the vectors ${\langle \sqrt{3}, 1 \rangle}$ and ${\langle \sin(x), \cos(x) \rangle}$ are proportional. This gives us a system of equations:

${\sin(x) = k\sqrt{3}}$

${\cos(x) = k}$

Now we can square each of these equations:

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

${\cos^2(x) = k^2}$

Adding these two equations, we get:

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

Using the Pythagorean identity ${\sin^2(x) + \cos^2(x) = 1}$, we have:

${1 = 4k^2}$

Therefore,

${k^2 = \frac{1}{4}}$

So, ${k = \pm \frac{1}{2}}$. We have to check both values of k to find the solutions.

If ${k = \frac{1}{2}}$, then ${\cos(x) = \frac{1}{2}}$ and ${\sin(x) = \frac{\sqrt{3}}{2}}$. This corresponds to the angle ${x = \frac{\pi}{3}}$. We've already checked this solution and it works!

If ${k = -\frac{1}{2}}$, then ${\cos(x) = -\frac{1}{2}}$ and ${\sin(x) = -\frac{\sqrt{3}}{2}}$. This corresponds to the angle ${x = \frac{4\pi}{3}}$. Let's substitute this into the original equation:

${\sqrt{3}\sin(\frac{4\pi}{3}) + \cos(\frac{4\pi}{3}) - 2 = \sqrt{3} \cdot (-\frac{\sqrt{3}}{2}) - \frac{1}{2} - 2 = -\frac{3}{2} - \frac{1}{2} - 2 = -4}$

This doesn't work! So, we discard this solution.

Conclusion

So, guys, using either algebraic manipulation and trigonometric identities, or the Cauchy-Schwarz Inequality, we've found that the only solution to the equation ${\sqrt{3}\sin(x) + \cos(x) - 2 = 0}$ in the interval ${[0, 2\pi)}$ is ${x = \frac{\pi}{3}}$. Remember to always check your solutions when you square equations, to avoid extraneous solutions! The general solution is ${x = \frac{\pi}{3} + 2n\pi}$, where n is an integer. I hope this helped, and if you have any questions, feel free to ask! Keep practicing, and you'll become a trigonometry pro in no time!