Solve $\sqrt{2} \cos X - 1 = 0$ For $0 \leq X \leq 2 \pi$

Hey there, math whizzes! Ever feel like you're staring down a trigonometry equation and just don't know where to begin? Well, you've come to the right place, guys! Today, we're diving deep into a specific problem that might look a little intimidating at first glance: finding the exact solutions of $\sqrt{2} \cos x - 1 = 0$ within the range of $0 \leq x \leq 2 \pi$. This is a super common type of question in trigonometry, and once you get the hang of the steps, you'll be solving them like a pro. We're talking about radians here, so get ready to dust off those unit circle vibes!

Isolating the Trigonometric Function

Alright, first things first, let's get our equation into a form where we can actually work with it. Our starting point is $\sqrt{2} \cos x - 1 = 0$. The goal here is to isolate the $\cos x$ term. Think of it like peeling an onion, layer by layer. We want to get $\cos x$ all by its lonesome on one side of the equation. So, what's the first step? We need to move that '-1' over to the other side. We can do this by adding 1 to both sides of the equation. This gives us $\sqrt{2} \cos x = 1$. See? Already looking a bit friendlier, right? Now, $\cos x$ is still being multiplied by $\sqrt{2}$. To get $\cos x$ completely by itself, we need to divide both sides by $\sqrt{2}$. This operation yields $\cos x = \frac{1}{\sqrt{2}}$.

Now, you might be thinking, "Wait a minute, that denominator has a square root!" And you'd be absolutely right. In mathematics, we often like to rationalize the denominator, which just means getting rid of that square root in the bottom. To do this, we multiply both the numerator and the denominator by $\sqrt{2}$. So, $\frac{1}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}$. Therefore, our simplified equation is $\cos x = \frac{\sqrt{2}}{2}$. This is the core equation we need to solve. Understanding how to manipulate and simplify equations like this is a fundamental skill, and it’s the bedrock upon which all further steps in trigonometry are built. Don't underestimate the power of basic algebraic manipulation; it's your best friend when tackling these more complex problems. We've successfully transformed the initial, slightly cumbersome equation into a much more recognizable form, setting the stage for identifying the specific angles.

Identifying Solutions on the Unit Circle

So, we've got $\cos x = \frac{\sqrt{2}}{2}$. Now, we need to find the angles $x$ within the specified range of $0 \leq x \leq 2 \pi$ where the cosine value is $\frac{\sqrt{2}}{2}$. This is where our trusty unit circle comes into play, guys! Remember, the cosine of an angle on the unit circle corresponds to the x-coordinate of the point where the terminal side of the angle intersects the circle. We're looking for points on the unit circle where the x-coordinate is $\frac{\sqrt{2}}{2}$.

This value, $\frac{\sqrt{2}}{2}$, is a very special value in trigonometry. It's associated with some of the most fundamental angles. Can you recall which angle has a cosine of $\frac{\sqrt{2}}{2}$? If you're thinking of $\frac{\pi}{4}$ (or 45 degrees), you are absolutely spot on! So, one of our solutions is $x = \frac{\pi}{4}$. This is our reference angle or principal value. Now, the problem specifies that we need to find all solutions within the interval $0 \leq x \leq 2 \pi$. The cosine function is positive in Quadrant I and Quadrant IV. Since $\frac{\pi}{4}$ is in Quadrant I, we need to find the angle in Quadrant IV that has the same cosine value.

To find the angle in Quadrant IV, we can use the relationship between the reference angle and the angles in each quadrant. For Quadrant IV, the angle is given by $2 \pi$ minus the reference angle. So, the other solution will be $x = 2 \pi - \frac{\pi}{4}$. Calculating this, we get $x = \frac{8 \pi}{4} - \frac{\pi}{4} = \frac{7 \pi}{4}$. Therefore, the two exact solutions for $\cos x = \frac{\sqrt{2}}{2}$ in the interval $0 \leq x \leq 2 \pi$ are $x = \frac{\pi}{4}$ and $x = \frac{7 \pi}{4}$. These are the angles where the cosine value is positive and equals $\frac{\sqrt{2}}{2}$. Mastering the unit circle and the properties of trigonometric functions across different quadrants is absolutely crucial for accurately solving these types of problems. It's not just about memorizing values; it's about understanding the cyclical nature and symmetry of these functions.

Verifying the Solutions

It's always a good practice, especially when you're learning or dealing with potentially tricky problems, to verify your solutions. This means plugging the values you found back into the original equation to make sure they actually work. Our original equation was $\sqrt{2} \cos x - 1 = 0$, and we found two potential solutions: $x = \frac{\pi}{4}$ and $x = \frac{7 \pi}{4}$.

Let's test $x = \frac{\pi}{4}$ first. We know that $\cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$. Plugging this into the equation: $\sqrt{2} \left(\frac{\sqrt{2}}{2}\right) - 1$. Multiplying $\sqrt{2}$ by $\frac{\sqrt{2}}{2}$ gives us $\frac{(\sqrt{2})^2}{2} = \frac{2}{2} = 1$. So, the equation becomes $1 - 1$, which equals 0. Perfect! The first solution checks out.

Now, let's test $x = \frac{7 \pi}{4}$. We know that the cosine function has a period of $2 \pi$, and $\frac{7 \pi}{4}$ is in Quadrant IV. The reference angle for $\frac{7 \pi}{4}$ is $\frac{\pi}{4}$. Since cosine is positive in Quadrant IV, $\cos \left(\frac{7 \pi}{4}\right)$ is the same as $\cos \left(\frac{\pi}{4}\right)$, which is $\frac{\sqrt{2}}{2}$. Plugging this into the equation: $\sqrt{2} \left(\frac{\sqrt{2}}{2}\right) - 1$. Just like with the first solution, this simplifies to $\frac{2}{2} - 1 = 1 - 1 = 0$. Fantastic! The second solution also works.

Both $x = \frac{\pi}{4}$ and $x = \frac{7 \pi}{4}$ satisfy the original equation and fall within the specified range of $0 \leq x \leq 2 \pi$. This verification step is super important because it helps catch any errors in your calculations or in understanding the properties of the trigonometric functions. It gives you that extra confidence that you've got the right answers. Always double-check your work, especially with those pesky signs and quadrants!

Conclusion and Final Answer

So, after going through all the steps – isolating the cosine function, using the unit circle to find the angles, and verifying our results – we've arrived at the exact solutions for the equation $\sqrt{2} \cos x - 1 = 0$ in the interval $0 \leq x \leq 2 \pi$. We found that $\cos x = \frac{\sqrt{2}}{2}$, and the angles within the given range where this is true are $x = \frac{\pi}{4}$ and $x = \frac{7 \pi}{4}$.

These are the only two solutions because the cosine function completes one full cycle between 0 and $2 \pi$, and $\frac{\sqrt{2}}{2}$ is a positive value, meaning the solutions must lie in Quadrant I (where $\frac{\pi}{4}$ is) and Quadrant IV (where $\frac{7 \pi}{4}$ is). In other quadrants, cosine is negative, so those angles won't satisfy our equation. Thus, the complete set of solutions is $\left\{\frac{\pi}{4}, \frac{7 \pi}{4}\right\}$.

Looking back at the multiple-choice options provided:

A. $\frac{\pi}{4}$ B. $\frac{\pi}{4}$ and $\frac{7 \pi}{4}$ C. $\frac{3 \pi}{4}$ and $\frac{5 \pi}{4}$ D. $\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \frac{7 \pi}{4}$

Our findings perfectly match option B. You guys absolutely crushed it! Remember these steps for your next trig problem: simplify, use the unit circle, consider all relevant quadrants, and always verify. Keep practicing, and you'll become trig masters in no time!