Solving Trigonometric Equations: A Step-by-Step Guide

Hey math enthusiasts! Let's dive into the fascinating world of trigonometry and explore how to solve some interesting equations. We'll be finding general solutions and, where necessary, rounding our answers to two decimal places. Buckle up, because we're about to have some fun with angles, sines, cosines, and more!

a) Solving 2cos(2x) + 1 = 0

Alright guys, let's start with our first equation: 2cos(2x) + 1 = 0. Our goal here is to isolate x. It's all about using our knowledge of trigonometric identities and algebraic manipulations to find the values of x that satisfy this equation. Trust me, it's easier than it sounds! First things first, we'll want to isolate the cosine term. We can do this by subtracting 1 from both sides of the equation, which gives us 2cos(2x) = -1. Next, divide both sides by 2, and we get cos(2x) = -1/2. Now, we need to find the angles whose cosine is -1/2. Remember the unit circle? The cosine function corresponds to the x-coordinate of a point on the unit circle. So, we're looking for angles where the x-coordinate is -1/2. We know that the cosine is negative in the second and third quadrants. The reference angle for which cos(θ) = 1/2 is 60° (or π/3 radians). Therefore, the angles in the second and third quadrants are 180° - 60° = 120° and 180° + 60° = 240°. But hold on, the argument of our cosine function is 2x, not x. So, we have 2x = 120° and 2x = 240°. To find the general solution, we need to consider all possible rotations around the unit circle. Cosine has a period of 360°, so we add multiples of 360° to our solutions. Thus, the general solutions for 2x are 120° + 360°n and 240° + 360°n, where 'n' is an integer. Finally, to find the general solutions for x, we divide these by 2: x = (120° + 360°n) / 2 = 60° + 180°n and x = (240° + 360°n) / 2 = 120° + 180°n. Since both solutions have a period of 180°, we can merge them into a single general solution: x = 60° + 180°n, where 'n' is an integer. Rounding to two decimal places, we get x = 60.00° + 180.00°n.

Now, let's think about this a bit more. The solutions are 60°, 240°, 420°, 600°, and so on (for n = 0, 1, 2, 3, etc.). These solutions repeat every 180 degrees. This is because cosine of a double angle is involved. You see how easy it is? Each step builds on the previous one. We started with a basic equation, isolated the cosine term, used our knowledge of the unit circle to find the angles, and then generalized the solution using the period of the cosine function. It is important to remember the unit circle. It is your friend!

b) Solving sin(x) = 3cos(x) for x ∈ [90°; 360°]

Alright, let's move on to something a little different. We have sin(x) = 3cos(x), and this time, we're looking for solutions within the interval [90°; 360°]. This means we only want the solutions that fall between 90 degrees and 360 degrees. Let's get to work! To solve this, a clever move is to divide both sides by cos(x), assuming cos(x) is not zero. This transforms the equation into something much more manageable. Dividing by cos(x) gives us sin(x) / cos(x) = 3, and since sin(x) / cos(x) is equal to tan(x), we can rewrite this as tan(x) = 3. Now, we need to find the angles whose tangent is 3. Use your calculator! Take the inverse tangent (arctan or tan⁻¹) of 3, and you'll get approximately 71.57°. This is the principal value, which falls in the first quadrant. However, the tangent function also has a positive value in the third quadrant. Remember, tangent is positive where both sine and cosine have the same sign (positive in the first and third quadrants, negative in the second and fourth). To find the angle in the third quadrant, we add 180° to the principal value: 71.57° + 180° = 251.57°. Therefore, the general solutions for tan(x) = 3 are x = 71.57° + 180°n, where 'n' is an integer. However, we're restricted to the interval [90°; 360°]. Looking at our general solution, only the value when n = 1 satisfies this condition. So, if we substitute n = 1 into our general solution, we get 71.57° + 180°(1) = 251.57°. So, the solution that satisfies the condition for x ∈ [90°; 360°] is approximately x = 251.57°. We can also see this graphically. The tangent function has a period of 180 degrees. The first solution is 71.57 degrees, and the next one is 71.57 + 180 = 251.57, which is within our specified range. The next value would be beyond our range.

So, we've successfully navigated this equation too! We used the handy trick of dividing by cosine, transformed the equation into a tangent function, found the principal value, considered all possible quadrants, and then applied our interval restriction to determine the final solution. The key takeaways here are the use of trigonometric identities and the importance of understanding the unit circle and the behavior of trigonometric functions.

c) Solving sin(x) = cos(3x)

Let's get even more creative with sin(x) = cos(3x)! This one may look a little tricky at first, but fear not. We will break it down step-by-step. The key to solving this equation is to recognize that we can express cosine in terms of sine using a trigonometric identity. Specifically, we know that cos(θ) = sin(90° - θ). We can use this to rewrite our equation. We can rewrite cos(3x) as sin(90° - 3x). This changes our equation to: sin(x) = sin(90° - 3x). Now, since the sines of the angles are equal, either the angles are equal, or they differ by multiples of 360° (or 2π radians). Therefore, one possibility is that: x = 90° - 3x + 360°n, where 'n' is an integer. Let's solve this for x. Adding 3x to both sides, we get 4x = 90° + 360°n. Dividing by 4, we find x = 22.5° + 90°n. Another possibility is that the angles are supplementary, or that they add up to 180°. So another possibility is: x = 180° - (90° - 3x) + 360°n. Simplifying that, we get x = 180° - 90° + 3x + 360°n, which becomes x = 90° + 3x + 360°n. Let's solve for x again: subtract 3x from both sides, -2x = 90° + 360°n. Divide both sides by -2: x = -45° - 180°n. Combining these two solutions, we have the general solutions for x. The first solution is x = 22.5° + 90°n. The second solution is x = -45° - 180°n. Notice that, using the period of the functions, these solutions can be simplified to x = 22.50° + 90.00°n and x = -45.00° - 180.00°n. These equations tell us the general solutions for x. In general, we have to consider all the solutions, or else the solution is not general.

d) Solving 6 - 10cos(x) = 3sin²(x) for x ∈ [-360°; 360°]

Alright, let's tackle the equation 6 - 10cos(x) = 3sin²(x), with the added constraint that x is in the interval [-360°; 360°]. This one involves a quadratic element due to the sin²(x) term, so we're going to need to use our identities to bring it into something more manageable. Our goal is to transform this equation into something we can solve for x. The first step is to recognize the Pythagorean identity: sin²(x) + cos²(x) = 1. We can rearrange this to get sin²(x) = 1 - cos²(x). Now, let's substitute this into our original equation: 6 - 10cos(x) = 3(1 - cos²(x)). Expanding the right side gives us 6 - 10cos(x) = 3 - 3cos²(x). Let's rearrange this to form a quadratic equation in terms of cos(x). Add 3cos²(x) to both sides, add 10cos(x) to both sides, and subtract 6 from both sides, and we get: 3cos²(x) - 10cos(x) - 3 = 0. This is a quadratic equation! Let's treat cos(x) as a variable. We can factor this: (3cos(x) + 1)(cos(x) - 3) = 0. Now, we have two possible equations: 3cos(x) + 1 = 0 and cos(x) - 3 = 0. Let's solve each one. For the first equation, 3cos(x) + 1 = 0, we can subtract 1 from both sides, then divide by 3: cos(x) = -1/3. Using a calculator, the inverse cosine of -1/3 (arccos or cos⁻¹) is approximately 109.47°. However, cosine is also negative in the third quadrant. So, the other solution is 360° - 109.47° = 250.53°. Our general solution will be these two solutions plus multiples of 360. Our first set of solutions: x = 109.47° + 360°n and x = 250.53° + 360°n, where 'n' is an integer. Let's think about the range. Using n=0, we get our two principal solutions. Using n=-1, we subtract 360 degrees from our principle solutions to obtain -250.53 and -109.47 degrees. This is still in our range of [-360, 360]. What happens when n=1? Then it falls outside our range, since you would be adding 360 degrees to the principle solutions. Let's deal with the second solution: cos(x) - 3 = 0. This means cos(x) = 3. However, the range of cosine is from -1 to 1. So, this equation has no solution. So finally, the solutions for x in the interval [-360°; 360°] are approximately x = -250.53°, -109.47°, 109.47°, and 250.53°.

There you have it! We've successfully solved this equation. We used the Pythagorean identity to rewrite the equation in terms of cosine, factored the resulting quadratic, and solved for x. Remember, understanding your trigonometric identities is key!

e) Solving 2 - sin(x)cos(x) - 3cos(2x) = 0

Finally, let's solve 2 - sin(x)cos(x) - 3cos(2x) = 0. This looks like another fun challenge! In this equation, we have a product of sine and cosine, and we have a cosine of a double angle. To solve this, we can try to use some identities to simplify the equation. The key here is to use double-angle identities. We can rewrite the equation using the double-angle identity for sine and cosine. Let's start with the double-angle identity for sine: sin(2x) = 2sin(x)cos(x). From this, we can deduce sin(x)cos(x) = (1/2)sin(2x). We'll also use the double-angle identity for cosine: cos(2x) = cos²(x) - sin²(x). We can rewrite the original equation as: 2 - (1/2)sin(2x) - 3cos(2x) = 0. Let's rearrange this to group the sine and cosine terms: 4 - sin(2x) - 6cos(2x) = 0. This is equal to sin(x)cos(x), so we can substitute that back in and rewrite as 2 - (1/2)sin(2x) - 3cos(2x) = 0. We can rearrange this to isolate terms containing trigonometric functions, which can become 4 - sin(2x) - 6cos(2x) = 0. However, this is quite complex and doesn't lead to a simple solution. Instead, let's explore another approach by recalling the double angle identity for cos(2x), which is cos(2x) = 1 - 2sin²(x). Substituting this into the original equation, we get: 2 - sin(x)cos(x) - 3(1 - 2sin²(x)) = 0. Simplifying this, we get 2 - sin(x)cos(x) - 3 + 6sin²(x) = 0. Now the equation will be: -1 - sin(x)cos(x) + 6sin²(x) = 0. If we use sin(2x) = 2sin(x)cos(x), we can substitute sin(x)cos(x) = (1/2)sin(2x). This changes to: -1 - (1/2)sin(2x) + 6sin²(x) = 0. This seems to be not very simple. Let's try another approach. Back to the equation: 2 - sin(x)cos(x) - 3cos(2x) = 0. We can rewrite sin(x)cos(x) = 1/2 sin(2x). Using the identity cos(2x) = cos²x - sin²x = 1 - 2sin²x, we can simplify the equation as follows: 2 - 1/2 sin(2x) - 3cos(2x) = 0. It is hard to solve analytically, so let's use another double angle identity: cos(2x) = 2cos²(x) - 1. We'll substitute this back into the equation: 2 - 1/2 sin(2x) - 3(2cos²(x) - 1) = 0. And that simplifies to 2 - 1/2 sin(2x) - 6cos²(x) + 3 = 0. This means: 5 - 1/2 sin(2x) - 6cos²(x) = 0. We can then substitute cos²(x) with 1 - sin²(x): 5 - 1/2 sin(2x) - 6(1 - sin²(x)) = 0. That's equal to: 5 - 1/2 sin(2x) - 6 + 6sin²(x) = 0. Which finally means: -1 - 1/2 sin(2x) + 6sin²(x) = 0. This can be expressed as: 12sin²x - sin2x - 2 = 0. You can factor to find: (4sinx + 1)(3sinx - 2) = 0. Therefore sinx = -1/4 or sinx = 2/3. For sinx = -1/4, we can say that x = arcsin(-1/4) = -14.48 degrees or x = 180 - arcsin(-1/4) = 194.48. For sinx = 2/3, we can say that x = arcsin(2/3) = 41.81 degrees or x = 180 - arcsin(2/3) = 138.19. The general solution is x = -14.48° + 360°n, 194.48° + 360°n, 41.81° + 360°n, or 138.19° + 360°n, where 'n' is an integer.

Conclusion

Well done, everyone! We've successfully solved a variety of trigonometric equations. Remember, practice makes perfect. Keep working on these problems and you'll become a trigonometry master in no time! Keep exploring the exciting world of mathematics. Until next time, happy solving!