Solving Trigonometric Equations: A Detailed Guide

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Hey guys! Today, we're diving deep into the world of trigonometric equations. We'll be tackling a specific problem step-by-step, making sure you understand the ins and outs of solving these types of equations. Whether you're prepping for an exam or just brushing up on your math skills, this guide is for you. Let's get started!

Understanding the Problem

Before we jump into solving, let's break down the equation we're dealing with. We have:

a) 2cos(2Ο€ + 2x) - 2 - √8 sinx = √6 + √12 sinx

This might look intimidating at first, but don't worry! We'll simplify it. The goal here is to find all values of 'x' that make this equation true. We'll also need to find the roots within a specific interval.

b) Find all roots of this equation that belong to the interval [-Ο€/2, Ο€/2]

This part narrows down our solutions to a specific range, which makes our job a bit easier. So, let's roll up our sleeves and get into the solution!

Step-by-Step Solution

Part a: Solving the Equation

Okay, let's tackle the first part: solving the trigonometric equation. Here’s how we can break it down:

1. Simplify the equation: Our main goal here is to use trigonometric identities and algebraic manipulations to get the equation into a more manageable form. This often involves combining like terms, applying identities, and rearranging the equation so that it’s easier to work with. Let’s start by simplifying the cosine term. Remember that cos(2Ο€ + ΞΈ) = cos(ΞΈ), because adding 2Ο€ doesn't change the angle's position on the unit circle. This identity is crucial for simplifying trigonometric expressions and making the equation more solvable.

  • Using the identity cos(2Ο€ + ΞΈ) = cos(ΞΈ):
    • 2cos(2Ο€ + 2x) becomes 2cos(2x)

Now our equation looks like this:

  • 2cos(2x) - 2 - √8 sinx = √6 + √12 sinx

2. Further Simplification: We can simplify the equation further by dealing with the square roots and using trigonometric identities to handle the double angle. Simplifying these terms will help us get closer to a form where we can isolate trigonometric functions and eventually solve for x. It’s like cleaning up the equation so we can see the path to the solution more clearly.

  • Simplify √8 and √12:
    • √8 = 2√2
    • √12 = 2√3

Our equation now looks like:

  • 2cos(2x) - 2 - 2√2 sinx = √6 + 2√3 sinx

3. Use the double angle identity: This is a crucial step. The double angle identity for cosine allows us to express cos(2x) in terms of sin(x), which will help us to combine terms and solve the equation. This is a common technique in solving trigonometric equations, where converting to a single trigonometric function can simplify the problem significantly.

  • Replace cos(2x) with 1 - 2sinΒ²(x):
    • 2(1 - 2sinΒ²(x)) - 2 - 2√2 sinx = √6 + 2√3 sinx

Expanding, we get:

  • 2 - 4sinΒ²(x) - 2 - 2√2 sinx = √6 + 2√3 sinx

4. Rearrange the equation: Get all terms on one side to set the equation to zero. This is a standard algebraic technique that allows us to use methods like factoring or the quadratic formula to solve for the variable. Bringing everything to one side is a fundamental step in solving many types of equations, not just trigonometric ones.

  • -4sinΒ²(x) - 2√2 sinx = √6 + 2√3 sinx

Move everything to the left side:

  • 4sinΒ²(x) + 2√2 sinx + 2√3 sinx + √6 = 0

5. Combine like terms: Group the sine terms together to make the equation look cleaner. This makes it easier to see if there are any simplifications we can make and prepares the equation for the next steps, like factoring or applying the quadratic formula. Combining like terms is a basic yet important step in simplifying equations.

  • 4sinΒ²(x) + (2√2 + 2√3)sinx + √6 = 0

6. Substitution: To make the equation easier to handle, let's substitute sin(x) with a variable, say 't'. This transforms the trigonometric equation into a quadratic equation, which we know how to solve. Substitution is a powerful technique that simplifies complex equations by temporarily replacing a part of the equation with a single variable.

  • Let t = sin(x):
    • 4tΒ² + (2√2 + 2√3)t + √6 = 0

7. Solve the quadratic equation: Now we have a standard quadratic equation. We can use the quadratic formula to find the values of 't'. This formula is a universal tool for solving quadratic equations, and it will give us the possible values for our substituted variable, t. From there, we can find the corresponding values for sin(x) and then solve for x.

  • The quadratic formula is:
    • t = (-b Β± √(bΒ² - 4ac)) / (2a)

Where: * a = 4 * b = 2√2 + 2√3 * c = √6

Plug in the values and calculate:

  • t = (-(2√2 + 2√3) Β± √((2√2 + 2√3)Β² - 4 * 4 * √6)) / (2 * 4)

Simplify the expression under the square root:

  • (2√2 + 2√3)Β² = 4 * 2 + 8√6 + 4 * 3 = 8 + 8√6 + 12 = 20 + 8√6
  • 4 * 4 * √6 = 16√6
  • 20 + 8√6 - 16√6 = 20 - 8√6

So, our equation becomes:

  • t = (-(2√2 + 2√3) Β± √(20 - 8√6)) / 8

8. Simplify the roots: The expression √(20 - 8√6) can be simplified further. This involves recognizing that the expression under the square root might be a perfect square in disguise. Simplifying square roots often makes the values more manageable and easier to work with in subsequent calculations.

  • Notice that 20 - 8√6 can be written as (2√3 - 2√2)Β²:
    • (2√3 - 2√2)Β² = (2√3)Β² - 2 * 2√3 * 2√2 + (2√2)Β² = 12 - 8√6 + 8 = 20 - 8√6

So, √(20 - 8√6) = |2√3 - 2√2| = 2√3 - 2√2 (since 2√3 > 2√2)

Now we have:

  • t = (-(2√2 + 2√3) Β± (2√3 - 2√2)) / 8

9. Calculate the two possible values for t: We have two cases to consider, one with the plus sign and one with the minus sign. This will give us the two possible values for t, which correspond to the two solutions of our quadratic equation. Each solution will lead to potential solutions for our original trigonometric equation.

  • Case 1 (using the plus sign):
    • t₁ = (-(2√2 + 2√3) + (2√3 - 2√2)) / 8 = (-4√2) / 8 = -√2 / 2
  • Case 2 (using the minus sign):
    • tβ‚‚ = (-(2√2 + 2√3) - (2√3 - 2√2)) / 8 = (-4√3) / 8 = -√3 / 2

10. Solve for x: Now we substitute back sin(x) for t and solve for x in each case. This is the final step in solving the equation, where we find the angles x that satisfy the original equation. Solving for x involves using inverse trigonometric functions and considering all possible solutions within the relevant range.

*   **Case 1: sin(x) = -√2 / 2**
    *   The solutions for x are x = -Ο€/4 + 2Ο€k and x = -3Ο€/4 + 2Ο€k, where k is an integer.
*   **Case 2: sin(x) = -√3 / 2**
    *   The solutions for x are x = -Ο€/3 + 2Ο€k and x = -2Ο€/3 + 2Ο€k, where k is an integer.

Part b: Finding Roots in the Interval [-Ο€/2, Ο€/2]

Now, let's find the roots that fall within the interval [-Ο€/2, Ο€/2]. This step involves checking the general solutions we found in part a and identifying which ones lie within the specified interval. We do this by plugging in different integer values for k and seeing if the resulting x values are within the range.

1. Check the solutions for sin(x) = -√2 / 2: We need to find the values of k that give us solutions in the interval [-Ο€/2, Ο€/2]. This involves simple arithmetic and understanding the range of values that trigonometric functions can take within a given interval.

  • x = -Ο€/4 + 2Ο€k
    • For k = 0, x = -Ο€/4 (which is in the interval)
    • For k = 1, x = -Ο€/4 + 2Ο€ = 7Ο€/4 (which is not in the interval)
    • For k = -1, x = -Ο€/4 - 2Ο€ = -9Ο€/4 (which is not in the interval)
  • x = -3Ο€/4 + 2Ο€k
    • For k = 0, x = -3Ο€/4 (which is not in the interval)
    • For k = 1, x = -3Ο€/4 + 2Ο€ = 5Ο€/4 (which is not in the interval)
    • For k = 1, x = -3Ο€/4 + 2Ο€ = 5Ο€/4 (not in the interval)

2. Check the solutions for sin(x) = -√3 / 2: Similarly, we check these solutions for values within our interval. This is a repetitive process but crucial for ensuring we only include the correct roots in our final answer.

  • x = -Ο€/3 + 2Ο€k
    • For k = 0, x = -Ο€/3 (which is in the interval)
    • For k = 1, x = -Ο€/3 + 2Ο€ = 5Ο€/3 (which is not in the interval)
    • For k = -1, x = -Ο€/3 - 2Ο€ = -7Ο€/3 (which is not in the interval)
  • x = -2Ο€/3 + 2Ο€k
    • For k = 0, x = -2Ο€/3 (which is not in the interval)
    • For k = 1, x = -2Ο€/3 + 2Ο€ = 4Ο€/3 (which is not in the interval)
    • For k = -1, x = -2Ο€/3 - 2Ο€ = -8Ο€/3 (which is not in the interval)

3. List the roots in the interval [-Ο€/2, Ο€/2]: Now, we compile the solutions that fall within the given interval. This is the final step in identifying the specific roots that satisfy both the equation and the interval condition. It’s like the grand finale where we present our findings.

  • The roots in the interval [-Ο€/2, Ο€/2] are:
    • x = -Ο€/4
    • x = -Ο€/3

Final Answer

So, to wrap it all up:

  • a) The general solutions to the equation are:
    • x = -Ο€/4 + 2Ο€k
    • x = -3Ο€/4 + 2Ο€k
    • x = -Ο€/3 + 2Ο€k
    • x = -2Ο€/3 + 2Ο€k
  • b) The roots in the interval [-Ο€/2, Ο€/2] are:
    • x = -Ο€/4
    • x = -Ο€/3

Key Takeaways

Solving trigonometric equations can seem daunting, but by breaking it down step-by-step, it becomes much more manageable. Remember these key points:

  1. Simplify: Use trigonometric identities to simplify the equation.
  2. Solve: Use algebraic techniques to isolate trigonometric functions.
  3. Substitute: Use substitution to convert to a more familiar form (like a quadratic equation).
  4. Find Roots: Identify the solutions within the given interval.

Keep practicing, and you'll become a pro at solving trigonometric equations in no time! You got this! Remember, the journey of a thousand miles begins with a single step, and in math, that single step is often simplification. Good luck, and happy solving!