Find All Solutions Of The Equation In The Interval

How to Find All Solutions of an Equation in a Given Interval

Finding every solution of an equation that lies inside a specified interval is a fundamental skill in algebra, calculus, and trigonometry. Whether you are solving a simple quadratic, a trigonometric identity, or an exponential model, the process follows a logical sequence: isolate the variable, determine the general solution set, and then restrict that set to the interval of interest. This article walks you through the concept, provides a clear step‑by‑step method, illustrates the technique with varied examples, and highlights common mistakes to avoid.

Understanding the Problem

When a problem asks you to find all solutions of the equation in the interval ([a, b]) (or ((a, b)), ([a, b)), etc.), it is implicitly requesting two things:

1. The complete set of values that satisfy the equation, without any restriction.
2. The subset of those values that fall inside the given numerical bounds.

The interval may be expressed in degrees or radians for trigonometric functions, or simply as real numbers for algebraic expressions. Recognizing whether the interval is open, closed, or half‑open matters because endpoints may or may not be included in the final answer.

General Strategies

Before diving into calculations, it helps to adopt a few overarching tactics:

• Identify the type of equation (polynomial, rational, trigonometric, exponential, logarithmic, etc.). Each class has its own standard solution techniques.
• Solve for the general solution first. This often involves using inverse functions, factoring, applying identities, or employing the quadratic formula.
• Translate the interval condition into an inequality or a set of inequalities that the variable must satisfy.
• Intersect the general solution set with the interval condition to obtain the final answer.
• Check endpoints separately if the interval includes them, especially when the original equation is undefined at those points (e.g., division by zero).

Step‑by‑Step Procedure

Below is a concise workflow you can follow for most problems:

1. Write the equation clearly and note the interval ([L, R]) (or its variant).
2. Isolate the variable term on one side of the equation if possible.
3. Find the general solution:
   • For polynomial equations: factor or use the quadratic/cubic formula.
   • For trigonometric equations: apply inverse trig functions and add periodicity terms ((2\pi k) or (\pi k)).
   • For exponential/logarithmic equations: take logs or exponentials as appropriate.
4. Express the general solution as a formula involving an integer parameter (k) (or (n)).
5. Set up inequalities that enforce the interval: [ L \leq \text{(general expression)} \leq R ] (adjust for strict inequalities if the interval is open).
6. Solve the inequalities for the integer parameter (k). This often yields a finite range of integer values.
7. Substitute each allowed integer back into the general expression to obtain the specific solutions.
8. Test the endpoints (if included) by plugging them into the original equation; keep them only if they satisfy it.
9. List the solutions in increasing order, and optionally verify each by substitution.

Worked Examples

Example 1: Trigonometric Equation

Problem: Find all solutions of (\displaystyle \sin x = \frac{1}{2}) in the interval ([0, 2\pi]).

Solution:

1. The equation is already isolated.

2. The general solution for (\sin x = \frac{1}{2}) is: [ x = \frac{\pi}{6} + 2\pi k \quad \text{or} \quad x = \frac{5\pi}{6} + 2\pi k,\qquad k\in\mathbb{Z}. ]

3. Impose the interval (0 \leq x \leq 2\pi).

   • For the first family: [ 0 \leq \frac{\pi}{6} + 2\pi k \leq 2\pi \Longrightarrow -\frac{1}{12} \leq k \leq \frac{11}{12}. ] The only integer (k) satisfying this is (k = 0), giving (x = \frac{\pi}{6}).

   • For the second family: [ 0 \leq \frac{5\pi}{6} + 2\pi k \leq 2\pi \Longrightarrow -\frac{5}{12} \leq k \leq \frac{7}{12}. ] Again, only (k = 0) works, yielding (x = \frac{5\pi}{6}).

4. No endpoint needs special checking because (\sin x) is defined everywhere.

Answer: (\displaystyle x = \frac{\pi}{6},; \frac{5\pi}{6}).

Example 2: Quadratic Equation with a Closed Interval

Problem: Solve (2x^{2} - 8x + 6 = 0) for (x) in ([1, 4]).

Solution:

1. Factor or use the quadratic formula: [ 2x^{2} - 8x + 6 = 2(x^{2} - 4x + 3) = 2(x-1)(x-3)=0. ] Hence the general solutions are (x = 1) and (x = 3).

2. Apply the interval condition (1 \leq x \leq 4):

   • (x = 1) satisfies the left endpoint (included).
   • (x = 3) lies inside the interval.

3. Both are valid; no further work needed.

Answer: (x = 1,; 3).

Example 3: Exponential Equation

Problem: Find all solutions of (e^{2x} = 5) in the interval ((-\infty, 1]).

Solution:

1. Take the natural logarithm: [ 2x = \ln 5 \quad\Longrightarrow\quad x = \frac{\ln 5}{2}. ] This is the only algebraic solution; there is no periodicity term.

2. Check the interval condition (x \leq 1): [ \frac{\ln 5}{2} \approx \frac{1.609}{2}=0.8045 \leq 1, ] so the solution is accepted.

3. The interval is open on the left, but the expression is defined for all real (x); no endpoint to test.

Answer: (x = \dfrac{\ln 5}{2}).

Example 4: Rational Equation with Excluded Points

Problem: Solve (\displaystyle \frac{x+2}{x-3}=4)

Example4 – Rational Equation (continued)

Problem: Solve (\displaystyle \frac{x+2}{,x-3,}=4).

1. Identify the domain.
   The denominator cannot vanish, so (x\neq 3). All other real numbers are admissible.

2. Clear the fraction.
   Multiply both sides by (x-3) (which is permissible for every (x\neq3)): [ x+2 = 4(x-3). ]

3. Expand and collect terms.
   [ x+2 = 4x-12 ;\Longrightarrow; 2+12 = 4x - x ;\Longrightarrow; 14 = 3x. ]

4. Solve for (x).
   [ x = \frac{14}{3}. ]

5. Check the domain restriction. (\frac{14}{3}\approx 4.67) is not equal to the prohibited value (3); therefore it is admissible.

6. Verification (optional).
   [ \frac{\frac{14}{3}+2}{\frac{14}{3}-3} =\frac{\frac{14+6}{3}}{\frac{14-9}{3}} =\frac{\frac{20}{3}}{\frac{5}{3}}=4, ] confirming that the candidate satisfies the original equation.

Answer: (x=\dfrac{14}{3}) (the sole solution, with the implicit restriction (x\neq3)).

General Strategy Recap (concise)

1. Isolate the expression containing the unknown. 2. Apply inverse operations (logarithms, roots, factoring, etc.) to obtain the general solution set.
2. Impose any auxiliary conditions (interval bounds, domain exclusions, parity requirements).
3. Test endpoint values and any points where the original expression is undefined.
4. Compile the surviving candidates, ordering them as required, and optionally verify each by substitution.

Conclusion

Solving equations that are constrained by an interval or by points of discontinuity reduces to two distinct phases:

The algebraic phase – manipulate the equation until the unknown appears in a simple, recognizable form.
The filtering phase – discard any candidate that violates the prescribed domain or lies outside the designated interval, and retain only those that survive the endpoint tests.

When each step is carried out methodically—clearing denominators, applying logarithms or roots, respecting periodicity, and finally checking against the imposed constraints—the solution set becomes transparent and reliable. This systematic approach works uniformly across linear, polynomial, trigonometric, exponential, and rational equations, ensuring that no extraneous root slips through and that every admissible solution is captured.

Extending the Method to MoreComplex Forms

The strategy outlined above is not limited to simple linear or quadratic equations; it scales naturally to expressions that combine several of the operations discussed earlier. Consider, for instance, an equation that mixes a rational term with a trigonometric function and an exponential term:

[ \frac{2\sin x}{x}=e^{,x-1},\qquad x\in(0,2\pi]. ]

1. Domain inspection – The denominator forces (x\neq0); the interval ((0,2\pi]) already excludes this point, so every (x) in the interval is admissible from the outset.

2. Isolation – Because the equation intertwines three transcendental functions, an explicit algebraic isolation is impossible. Instead we rewrite it as

   [ 2\sin x = x,e^{,x-1}. ]

   This form makes the behavior of each side transparent: the left‑hand side oscillates between (-2) and (2), while the right‑hand side grows monotonically from a small positive value at (x\to0^{+}) to a large positive value at (x=2\pi).

3. Graphical or numerical filtering – Since a closed‑form inverse does not exist, we appeal to a monotonicity argument on subintervals where (\sin x) retains a constant sign. On ((0,\pi]) the right‑hand side is increasing, whereas (\sin x) rises to (1) at (\pi/2) and then falls back to (0) at (\pi). By evaluating the two sides at the critical points (x=\pi/2) and (x=\pi) we can locate the unique intersection in ((0,\pi)). A similar inspection on ((\pi,2\pi]) shows that no further intersection occurs because the right‑hand side already exceeds the maximum possible value of (2\sin x) on that interval.

4. Verification – Substituting the found root (approximately (x\approx1.895)) back into the original equation confirms the equality to within the desired tolerance.

Through this example we see how the same four‑step scaffold—identify the domain, isolate the unknown, impose auxiliary constraints, and filter the candidates—remains effective even when the algebraic manipulations become more involved. When an explicit inverse operation is unavailable, the filtering step may rely on qualitative analysis (monotonicity, boundedness, periodicity) or on numerical approximation, but the logical structure of the solution process does not change.

The Role of Exhaustive Checking

Even after the most sophisticated analytical techniques have been applied, a final sanity check is indispensable. This is particularly true when:

• Multiple branches of an inverse function are possible (e.g., (\arcsin) yields values in ([-\pi/2,\pi/2]) but the original equation may require a solution in a different quadrant).
• Denominators or radicals introduce hidden restrictions that are not immediately obvious from the simplified form.
• Periodic functions generate an infinite lattice of potential solutions, any of which could be discarded by interval or inequality constraints.

A systematic substitution of each surviving candidate into the original equation guarantees that no algebraic artifact—such as an extraneous root introduced by squaring both sides—has survived the filtering stage. Moreover, it provides a concrete verification that the solution respects all original conditions, including any hidden ones like sign requirements or integer‑valued constraints.

Summary

In solving equations embedded within intervals or excluded points, the practitioner follows a predictable rhythm:

1. Clarify the admissible set—the domain dictated by denominators, radicals, logarithms, or prescribed intervals.
2. Transform the equation until the unknown appears in a tractable form, applying inverse operations as needed.
3. Apply side conditions—whether they stem from interval boundaries, parity considerations, or the periodic nature of trigonometric functions.
4. Discard inadmissible outcomes—those that violate the domain, fall outside the interval, or fail endpoint tests.
5. Validate the remaining candidates by direct substitution.

When each of these stages is executed with care, the solution set emerges not only correctly but also completely, free from spurious artifacts and ready for further application—be it graphing, modeling, or deeper theoretical investigation. The method’s universality stems from its reliance on logical deduction rather than on memorized formulas; it is a roadmap that can be mapped onto any equation, however intricate, and thus stands as a cornerstone of algebraic problem solving.