Solving Trig Equations: A Step-by-Step Guide

Hey math enthusiasts! Let's dive into some interesting trigonometric equations. We'll break down each problem step-by-step, making sure everyone understands the concepts. Get ready to flex those math muscles!

Part 1: Solving 5cos²x = 4 + sin x

Alright guys, let's tackle the first problem: solving 5cos²x = 4 + sin x for 0° < x < 180°. This is a classic example of a trigonometric equation where we need to use some clever substitutions and algebraic manipulation. The key here is to get everything in terms of a single trigonometric function. Remember those trig identities? They're our best friends in this situation. Specifically, we'll use the identity cos²x + sin²x = 1. This allows us to rewrite the cosine term in terms of sine.

First, let's rearrange the identity to isolate cos²x: cos²x = 1 - sin²x. Now, we can substitute this into our original equation. So, 5cos²x = 4 + sin x becomes 5(1 - sin²x) = 4 + sin x. Let's expand this: 5 - 5sin²x = 4 + sin x. Now, let's rearrange everything to one side to get a quadratic equation: 0 = 5sin²x + sin x - 1. This is a quadratic equation in terms of sin x. To make it easier to solve, let's substitute y = sin x. This gives us 5y² + y - 1 = 0. We can solve this quadratic equation using the quadratic formula: y = (-b ± √(b² - 4ac)) / 2a. In our case, a = 5, b = 1, and c = -1. Plugging these values into the formula, we get y = (-1 ± √(1² - 45(-1))) / (2*5) which simplifies to y = (-1 ± √21) / 10. So, we have two possible values for y (and therefore sin x): y₁ = (-1 + √21) / 10 ≈ 0.358 and y₂ = (-1 - √21) / 10 ≈ -0.558. Remember that y = sin x, so we now need to find the values of x that correspond to these sine values. We are looking for solutions in the range 0° < x < 180°.

For y₁ ≈ 0.358, we find x₁ ≈ arcsin(0.358) ≈ 21°. For y₂ ≈ -0.558, since the sine function is negative in the third and fourth quadrants, and we're only looking between 0° and 180°, there is no solution here, as the values are outside of the allowed range. Therefore, the only solution to the original equation within the specified range is approximately 21°. Understanding these steps is crucial. We utilized trigonometric identities to simplify the equation, converted it into a more manageable quadratic form, and then applied the quadratic formula to solve for the unknown angle. Remember to always consider the specified range of the angle and discard any extraneous solutions that fall outside of it. This methodical approach is key to successfully solving trigonometric equations.

Part 2: Solving 5sin²θ = 4 + cos θ

Okay, let's move on to the second part: solving 5sin²θ = 4 + cos θ for 0° < θ < 180°. This problem is quite similar to the first one, but this time we're dealing with sine and cosine directly. The strategy remains the same: use the trigonometric identity to express everything in terms of a single trigonometric function. In this case, we'll want to convert the sin²θ term using the identity sin²θ + cos²θ = 1. This gives us sin²θ = 1 - cos²θ. Now, substitute this into the equation: 5(1 - cos²θ) = 4 + cos θ. Expand this out: 5 - 5cos²θ = 4 + cos θ. Rearrange the terms to get a quadratic equation in terms of cos θ: 0 = 5cos²θ + cos θ - 1. Just like before, let's make a substitution to simplify things. Let's say z = cos θ. The equation becomes 5z² + z - 1 = 0. This is the same quadratic equation we saw in the first problem! The solutions are exactly the same: z = (-1 ± √21) / 10. Therefore, cos θ ≈ 0.358 and cos θ ≈ -0.558. We need to find the values of θ in the range 0° < θ < 180°.

For cos θ ≈ 0.358, we find θ₁ ≈ arccos(0.358) ≈ 69°. For cos θ ≈ -0.558, we find θ₂ ≈ arccos(-0.558) ≈ 124°. Both of these angles fall within the specified range. Thus, the solutions for θ are approximately 69° and 124°. Notice the parallels between this problem and the previous one. We used the same trigonometric identity, the same algebraic manipulations, and even ended up with the same quadratic equation. This highlights the importance of recognizing patterns and applying the appropriate strategies to solve trigonometric equations effectively. Understanding the cyclical nature of trigonometric functions and the importance of considering the specified range are crucial. Pay close attention to the signs and quadrants to ensure that your solutions are valid within the given constraints. By consistently applying these principles, you will be well-equipped to tackle a wide variety of trigonometric problems.

Part 3: Solving 5cot²φ - 4cotφ = 0

Alright, let's jump into the third part: solving 5cot²φ - 4cotφ = 0 for 0 < φ < π. This equation looks a little different, but the approach remains familiar. We've got cotangent functions, so let's simplify by factoring. Notice that both terms have a common factor of cot φ. So, we can factor out cot φ: cot φ (5 cot φ - 4) = 0. This gives us two possible solutions: either cot φ = 0 or 5 cot φ - 4 = 0. Let's analyze each case.

First, if cot φ = 0, then φ must be the angle where the cotangent function is zero. Remember that cotangent is the reciprocal of the tangent function (cot φ = 1/tan φ or cot φ = cos φ/sin φ). The cotangent function is zero when the cosine is zero, and the sine is not zero. In the range 0 < φ < π (which is the same as 0° < φ < 180°), the only angle where this occurs is when φ = π/2 (90°). Second, let's solve 5 cot φ - 4 = 0. Rearranging gives us 5 cot φ = 4, so cot φ = 4/5. To find the angle φ, we take the inverse cotangent (arccot) of 4/5: φ = arccot(4/5). This will give us an angle in the range 0 < φ < π. The arccot(4/5) is approximately 0.675 radians (or about 36.87°). So, we have two solutions: φ₁ = π/2 and φ₂ ≈ 0.675 radians. These solutions both fall within the specified range of 0 < φ < π. This example showcases the power of factoring in solving trigonometric equations. By identifying a common factor, we were able to break down the equation into simpler components. Understanding the relationship between different trigonometric functions, like cotangent and tangent, is essential for simplifying and solving these equations. Remember to always check your solutions to ensure they fall within the specified range and that the trigonometric functions are defined for those angles. Practice, practice, practice! The more problems you solve, the more comfortable you'll become with recognizing patterns and applying the correct strategies.

I hope this comprehensive guide has helped you understand how to solve these trigonometric equations. Remember to practice and stay curious! Keep up the great work, everyone! And always, always double-check your answers!