Solving 3cos(A) - 1 = 0 Trigonometric Equation Between 0 And 360 Degrees

Hey guys! Today, we're diving into a trigonometric problem that many students find tricky but is actually pretty straightforward once you break it down. We're going to solve the equation 3cos(A) - 1 = 0 for all the angles A that fall between 0° and 360°. This means we're looking for solutions within a full circle. Let's get started and make sure we understand each step clearly.

Understanding the Problem

Before we jump into the solution, let's quickly recap what we're dealing with. We have a trigonometric equation involving the cosine function, and we need to find the angles that make this equation true. The cosine function, as you might remember, relates to the x-coordinate of a point on the unit circle. So, essentially, we're trying to find the angles where the x-coordinate on the unit circle satisfies our equation. It's crucial to have a solid grasp of the unit circle and cosine's behavior within it to tackle these problems effectively. Understanding the cosine function is pivotal here. The cosine function oscillates between -1 and 1. Visualizing the graph of cosine can also be incredibly helpful. It starts at 1 when the angle is 0°, goes down to 0 at 90°, reaches -1 at 180°, returns to 0 at 270°, and finally gets back to 1 at 360°. This cyclic nature means that cosine has the same value at multiple angles, which is why we often have more than one solution in the interval of 0° to 360°.

Isolating the Cosine Function

The first step in solving any equation is to isolate the variable we're interested in. In this case, our variable is cos(A). We want to get cos(A) by itself on one side of the equation. To do this, we'll start by adding 1 to both sides of the equation: 3cos(A) - 1 + 1 = 0 + 1, which simplifies to 3cos(A) = 1. Next, we'll divide both sides by 3 to completely isolate cos(A): 3cos(A) / 3 = 1 / 3, giving us cos(A) = 1/3. So, now we know that we're looking for angles A where the cosine is equal to 1/3. This is a positive value, which is important because it tells us which quadrants to focus on. Remember, cosine is positive in the first and fourth quadrants of the unit circle. Now that we've isolated cos(A), we have a clearer target. We know the value we want cosine to be, and we can proceed to find the angles that satisfy this condition.

Finding the Reference Angle

Now that we have cos(A) = 1/3, we need to find the angle A whose cosine is 1/3. This is where the inverse cosine function, often written as arccos or cos⁻¹, comes into play. Your calculator should have this function, usually accessed by pressing the "shift" or "second" key and then the cosine key. To find the reference angle, we calculate A = cos⁻¹(1/3). Make sure your calculator is in degree mode (not radians!) for this calculation. When you plug in cos⁻¹(1/3), you'll get an approximate value. This value is our reference angle, which we'll call A_ref. The reference angle is the acute angle formed between the terminal side of our angle and the x-axis. It's a crucial piece of information because it helps us find all the angles in the interval 0° to 360° that have the same cosine value. Using your calculator, you should find that A_ref is approximately 70.53°. Keep this value handy, as we'll use it to determine the solutions in the correct quadrants. The reference angle simplifies the process of finding all solutions because it provides a base angle that we can use to find angles in other quadrants.

Determining Solutions in the Correct Quadrants

As we discussed earlier, cosine is positive in the first and fourth quadrants. Our reference angle, A_ref ≈ 70.53°, is in the first quadrant, so that's one of our solutions. Let's call it A1. So, A1 = 70.53°. Now, we need to find the solution in the fourth quadrant. In the fourth quadrant, angles are measured clockwise from the positive x-axis. To find the angle in the fourth quadrant that has the same cosine value as A_ref, we subtract A_ref from 360°. This gives us A2 = 360° - A_ref. Plugging in our value for A_ref, we get A2 = 360° - 70.53° ≈ 289.47°. So, our two solutions in the interval 0° to 360° are approximately 70.53° and 289.47°. Remember, quadrant analysis is crucial in trigonometry. The sign of the trigonometric function in each quadrant dictates where the solutions lie. By understanding the signs of sine, cosine, and tangent in each quadrant, you can accurately determine all the solutions within the desired interval.

Checking the Solutions

It's always a good idea to check our solutions to make sure they're correct. We can do this by plugging our values for A back into the original equation, 3cos(A) - 1 = 0. Let's start with A1 = 70.53°: 3cos(70.53°) - 1 ≈ 3(0.333) - 1 ≈ 1 - 1 = 0. This checks out! Now, let's try A2 = 289.47°: 3cos(289.47°) - 1 ≈ 3(0.333) - 1 ≈ 1 - 1 = 0. This also checks out! So, we can be confident that our solutions are correct. Verifying solutions is a best practice in mathematics. It helps you catch any potential errors and ensures that your answers are accurate.

Final Answer and Options

Alright, we've found our solutions: A1 ≈ 70.53° and A2 ≈ 289.47°. Now, let's look at the answer options provided: a. A1=132° and A2=342° b. A1=48.7° and A2=375.3° c. A1=73.5° and A2=150° d. A1=70.52° and A2=289.47° Comparing our solutions to the options, we can see that option d, A1=70.52° and A2=289.47°, is the closest match. Note that there's a slight difference in the first solution (70.52° instead of 70.53°), but this is likely due to rounding. So, the correct answer is d. A1=70.52° and A2=289.47°. Choosing the correct option is the final step, but it's important to ensure that you've understood each step of the solution process to avoid errors. Remember, accuracy in calculations is essential for arriving at the correct answer.

Conclusion

So, there you have it! We've successfully solved the trigonometric equation 3cos(A) - 1 = 0 for values of A between 0° and 360°. We isolated the cosine function, found the reference angle, determined the solutions in the correct quadrants, and checked our answers. Remember, practice makes perfect, so keep working on these types of problems, and you'll become a trigonometry pro in no time! If you have any questions, feel free to ask. Keep up the great work, guys!