Sigma Percentile
JEE Main 2022 (25 July Shift 1)
LEVELJEE Main

Animated Solution for Mathematics - Trigonometry: The number of solutions of , such that is :

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Visualized Solution

The Problem Setup

  • Equation:
  • Domain:
  • Goal: Find the total number of solutions.

Visualizing the Right Hand Side

  • Let's plot the Right Hand Side (RHS) function.
  • This is the standard sine wave over the given domain.

Visualizing the Left Hand Side

  • Now, let's plot the Left Hand Side (LHS) function.
  • The absolute value flips the negative parts of the cosine wave above the x-axis.

The Hidden Constraint

  • Notice the absolute value on the LHS:
  • For the equation to hold, the RHS must also be non-negative.
  • Therefore, we must have

Locating Valid Regions

  • We only consider regions where the sine curve is above or on the x-axis.
  • Let's look at the first positive interval:

Solving in the Principal Interval

  • In the interval , we look for intersection points.
  • Graphically, the green curve and blue curve intersect twice.

Algebraic Confirmation

  • Algebraically,
  • In , the solutions are and
  • So, there are exactly solutions in this interval.

Finding Other Valid Intervals

  • Because trigonometric functions are periodic, this pattern repeats.
  • We identify all other intervals in where

Counting the Intervals

  • The valid intervals are:
  • Total number of valid intervals =

Generalizing the Solutions

  • Due to periodicity, each interval of length contains exactly solutions.
  • Let's mark the intersection points in all valid intervals.

Final Calculation

  • Total solutions = (Number of intervals) (Solutions per interval)
  • Total solutions =
  • Conclusion: There are solutions in the given domain.

The Sigma Insight: General Solution of Trigonometric Equations

Solution Diagram

Analyzing the Setup

We are tasked with solving the equation within the domain . This problem requires a careful balance of algebraic manipulation and geometric intuition.
The left-hand side involves an absolute value, which is inherently non-negative, i.e., . Consequently, for the equation to hold, the right-hand side must satisfy the constraint:
This constraint acts as our "Gatekeeper." We are restricted to the intervals where the sine function is non-negative, specifically within the quadrants where the sine wave resides above the -axis.

Visualizing the Periodic Rhythm

The sine function completes four full cycles within the interval . We must identify the specific sub-intervals where :
1. In , . 2. In , . 3. In , . 4. In , .
These four intervals represent the regions where potential solutions exist. The periodic nature of these functions ensures that the behavior within each interval is identical.

The Algebraic Solution

Let us examine the interval to determine the specific values of . We split the equation into two cases based on the sign of :
Case 1: . The equation simplifies to , which implies . Within , this yields:
Case 2: . The equation simplifies to , which implies . Within , this yields:
Thus, we have found exactly two solutions within the first valid interval. Because the functions and exhibit a periodic pattern, each of the four identified intervals will contribute exactly two solutions.

Final Calculation

We have identified valid intervals, and each interval provides distinct solutions. The total number of solutions is calculated as:
By respecting the constraints of the absolute value and leveraging the periodicity of the trigonometric functions, we have successfully navigated the domain. The total number of solutions for the given equation is 8.

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