Solve The Equation For Exact Solutions Over The Interval

Let's dive into the world of solving equations for exact solutions over a specified interval. This involves a combination of algebraic manipulation, trigonometric identities (if applicable), and careful consideration of the interval's boundaries.

Understanding the Problem

When we talk about "solving an equation," we generally mean finding the values of the variable(s) that make the equation true. These values are called the solutions or roots of the equation. However, when we add the phrase "for exact solutions over the interval," we're placing a constraint on the possible solutions. We're not just looking for any solution; we're only interested in solutions that fall within the given interval. This is particularly relevant for trigonometric equations, which often have infinitely many solutions due to the periodic nature of trigonometric functions.

Let's break down the key components:

• Equation: This is the mathematical statement expressing equality between two expressions. It could be algebraic (e.g., x² + 2x + 1 = 0), trigonometric (e.g., sin(x) = 0.5), exponential (e.g., 2^x = 8), or a combination of these.
• Exact Solutions: This implies that we want solutions expressed in their simplest, most precise form. This usually means avoiding decimal approximations. For example, expressing the solution as √2 instead of 1.414 is preferred. Similarly, solutions involving π are preferred in their symbolic form rather than as 3.14159.
• Interval: This is a range of values within which we are seeking solutions. It's typically written in interval notation, such as [a, b] (inclusive of both a and b), (a, b) (exclusive of both a and b), [a, b) (inclusive of a, exclusive of b), or (a, b] (exclusive of a, inclusive of b). The interval could be finite (e.g., [0, 2π]) or infinite (e.g., (-∞, ∞)), although the instruction of 'exact solutions' usually suggests a finite interval.

General Steps for Solving Equations Over an Interval

While the specific techniques will vary depending on the type of equation, here's a general outline of the process:

1. Solve the Equation: Use appropriate algebraic, trigonometric, or other mathematical techniques to find the general solution(s) of the equation. This step typically involves isolating the variable or using factoring, quadratic formula, trigonometric identities, or other relevant methods.
2. Identify Solutions within the Interval: Once you have the general solution(s), you need to determine which of those solutions fall within the specified interval. This often involves adding or subtracting multiples of the period (if applicable) to the general solution until you find the solutions that lie within the interval.
3. Express Solutions Exactly: Ensure your solutions are expressed in their simplest, most precise form, avoiding decimal approximations whenever possible.
4. Check Your Solutions: Substitute each solution back into the original equation to verify that it satisfies the equation. Also, double-check that each solution indeed falls within the specified interval.

Solving Algebraic Equations Over an Interval

Algebraic equations involve variables, constants, and mathematical operations like addition, subtraction, multiplication, division, and exponentiation. Let's look at some examples.

Example 1: Linear Equation

Solve the equation 3x + 5 = 14 for x over the interval [-2, 5].

1. Solve the Equation:

   • 3x + 5 = 14
   • 3x = 9
   • x = 3

2. Identify Solutions within the Interval:

   • The solution x = 3 falls within the interval [-2, 5].

3. Express Solutions Exactly:

   • The solution is already in its exact form: x = 3.

4. Check Your Solutions:

   • 3(3) + 5 = 9 + 5 = 14. The equation is satisfied.
   • 3 is within the interval [-2, 5].

Solution: x = 3

Example 2: Quadratic Equation

Solve the equation x² - 5x + 6 = 0 for x over the interval [0, 4].

1. Solve the Equation:

   • We can factor the quadratic equation: (x - 2)(x - 3) = 0
   • This gives us two possible solutions: x = 2 and x = 3.

2. Identify Solutions within the Interval:

   • Both x = 2 and x = 3 fall within the interval [0, 4].

3. Express Solutions Exactly:

   • The solutions are already in their exact form: x = 2 and x = 3.

4. Check Your Solutions:

   • For x = 2: 2² - 5(2) + 6 = 4 - 10 + 6 = 0. The equation is satisfied.
   • For x = 3: 3² - 5(3) + 6 = 9 - 15 + 6 = 0. The equation is satisfied.
   • Both 2 and 3 are within the interval [0, 4].

Solution: x = 2, x = 3

Example 3: Rational Equation

Solve the equation (x+1)/(x-2) = 3 for x over the interval [-1, 5).

1. Solve the Equation:

   • Multiply both sides by (x-2): x+1 = 3(x-2)
   • Simplify: x+1 = 3x - 6
   • Isolate x: 2x = 7
   • x = 7/2 = 3.5

2. Identify Solutions within the Interval:

   • 3.5 falls within the interval [-1, 5) since -1 <= 3.5 < 5.

3. Express Solutions Exactly:

   • The solution is x = 7/2

4. Check Your Solutions:

   • ((7/2)+1)/((7/2)-2) = (9/2)/(3/2) = 3. The equation is satisfied.
   • 7/2 is within the interval [-1, 5).

Solution: x = 7/2

Solving Trigonometric Equations Over an Interval

Trigonometric equations involve trigonometric functions like sine, cosine, tangent, cotangent, secant, and cosecant. Solving these equations requires knowledge of trigonometric identities and the periodic nature of these functions.

Example 1: Basic Sine Equation

Solve the equation sin(x) = 0.5 for x over the interval [0, 2π).

1. Solve the Equation:

   • We know that sin(π/6) = 0.5. So, one solution is x = π/6.
   • Since sin(x) = sin(π - x), another solution is x = π - π/6 = 5π/6.
   • The general solution for sin(x) = 0.5 is x = π/6 + 2πk or x = 5π/6 + 2πk, where k is an integer.

2. Identify Solutions within the Interval:

   • For k = 0, we have x = π/6 and x = 5π/6, both of which fall within the interval [0, 2π).
   • For k = 1, we have x = π/6 + 2π and x = 5π/6 + 2π, both of which are outside the interval [0, 2π).
   • For k = -1, we have x = π/6 - 2π and x = 5π/6 - 2π, both of which are outside the interval [0, 2π).

3. Express Solutions Exactly:

   • The solutions are x = π/6 and x = 5π/6.

4. Check Your Solutions:

   • sin(π/6) = 0.5
   • sin(5π/6) = 0.5
   • Both π/6 and 5π/6 are within the interval [0, 2π).

Solution: x = π/6, x = 5π/6

Example 2: Cosine Equation with a Multiple Angle

Solve the equation cos(2x) = √3/2 for x over the interval [0, π].

1. Solve the Equation:

   • We know that cos(π/6) = √3/2. So, 2x = π/6.
   • Also, cos(-π/6) = √3/2, so 2x = -π/6.
   • The general solution for cos(2x) = √3/2 is 2x = ±π/6 + 2πk, where k is an integer.
   • Dividing by 2, we get x = ±π/12 + πk.

2. Identify Solutions within the Interval:

   • For k = 0, we have x = π/12 and x = -π/12. Since we are looking for solutions over [0, π], we discard x = -π/12. x = π/12 is a valid solution.
   • For k = 1, we have x = π/12 + π = 13π/12 and x = -π/12 + π = 11π/12. Since we are looking for solutions over [0, π], we keep x = 11π/12 and discard x = 13π/12.
   • For k = 2, we have x = π/12 + 2π and x = -π/12 + 2π, both outside the interval [0, π].
   • For k = -1, we have x = π/12 - π and x = -π/12 - π, both outside the interval [0, π].

3. Express Solutions Exactly:

   • The solutions are x = π/12 and x = 11π/12.

4. Check Your Solutions:

   • cos(2*(π/12)) = cos(π/6) = √3/2
   • cos(2*(11π/12)) = cos(11π/6) = √3/2
   • Both π/12 and 11π/12 are within the interval [0, π].

Solution: x = π/12, x = 11π/12

Example 3: Tangent Equation

Solve the equation tan(x) = 1 for x over the interval (-π/2, π/2).

1. Solve the Equation:

   • We know that tan(π/4) = 1. The general solution for tan(x) = 1 is x = π/4 + kπ, where k is an integer.

2. Identify Solutions within the Interval:

   • For k = 0, x = π/4, which is within the interval (-π/2, π/2).
   • For k = 1, x = π/4 + π = 5π/4, which is outside the interval (-π/2, π/2).
   • For k = -1, x = π/4 - π = -3π/4, which is outside the interval (-π/2, π/2).

3. Express Solutions Exactly:

   • The solution is x = π/4.

4. Check Your Solutions:

   • tan(π/4) = 1
   • π/4 is within the interval (-π/2, π/2).

Solution: x = π/4

Solving Exponential and Logarithmic Equations Over an Interval

Exponential and logarithmic equations require understanding the properties of exponents and logarithms.

Example 1: Exponential Equation

Solve the equation 2^x = 8 for x over the interval [-1, 5].

1. Solve the Equation:

   • We know that 2³ = 8. So, x = 3.

2. Identify Solutions within the Interval:

   • The solution x = 3 falls within the interval [-1, 5].

3. Express Solutions Exactly:

   • The solution is already in its exact form: x = 3.

4. Check Your Solutions:

   • 2³ = 8. The equation is satisfied.
   • 3 is within the interval [-1, 5].

Solution: x = 3

Example 2: Logarithmic Equation

Solve the equation ln(x) = 0 for x over the interval (0, 10].

1. Solve the Equation:

   • We know that ln(1) = 0. So, x = 1.

2. Identify Solutions within the Interval:

   • The solution x = 1 falls within the interval (0, 10].

3. Express Solutions Exactly:

   • The solution is already in its exact form: x = 1.

4. Check Your Solutions:

   • ln(1) = 0. The equation is satisfied.
   • 1 is within the interval (0, 10].

Solution: x = 1

Example 3: Combining Logarithmic Properties

Solve the equation log₂(x + 2) + log₂(x - 1) = 2 for x over the interval (1, ∞).

1. Solve the Equation:

   • Using the logarithmic property log_a(b) + log_a(c) = log_a(bc), we have: log₂((x + 2)(x - 1)) = 2
   • Convert to exponential form: (x + 2)(x - 1) = 2² = 4
   • Expand and simplify: x² + x - 2 = 4
   • x² + x - 6 = 0
   • Factor: (x + 3)(x - 2) = 0
   • Therefore, x = -3 or x = 2

2. Identify Solutions within the Interval:

   • x = -3 is not in the interval (1, ∞), so we discard it.
   • x = 2 is in the interval (1, ∞).

3. Express Solutions Exactly:

   • The solution is x = 2

4. Check Your Solutions:

   • log₂(2 + 2) + log₂(2 - 1) = log₂(4) + log₂(1) = 2 + 0 = 2
   • x = 2 is within the interval (1, ∞).

Solution: x = 2

Important Considerations and Common Mistakes

• Periodicity of Trigonometric Functions: Always remember the periodic nature of trigonometric functions when finding solutions within an interval. Add or subtract multiples of the period (2π for sine and cosine, π for tangent) to the general solution to find all solutions within the given interval.
• Domain Restrictions: Be mindful of domain restrictions for functions like logarithms and rational expressions. For example, the argument of a logarithm must be positive, and the denominator of a rational expression cannot be zero. Discard any solutions that violate these restrictions.
• Interval Notation: Pay close attention to whether the interval is open, closed, or half-open. This determines whether the endpoints of the interval are included in the solution set.
• Extraneous Solutions: When solving equations involving radicals or rational expressions, it's crucial to check your solutions to ensure they are not extraneous (i.e., solutions that satisfy the transformed equation but not the original equation).
• Approximations: Unless explicitly instructed to provide approximate solutions, always aim for exact solutions. Avoid rounding off intermediate results, as this can lead to inaccuracies.

Conclusion

Solving equations for exact solutions over a specified interval requires a systematic approach and a solid understanding of algebraic, trigonometric, exponential, and logarithmic principles. By following the steps outlined above, paying attention to domain restrictions and periodicity, and carefully checking your solutions, you can confidently tackle these types of problems. Remember, practice is key to mastering these techniques. Working through a variety of examples will help you develop the skills and intuition needed to solve even the most challenging equations.