Conquer Complex Trig Equations: A JEE Mains Deep Dive

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Conquer Complex Trig Equations: A JEE Mains Deep Dive

Hey guys, ever stared at a math problem and thought, "Whoa, where do I even begin?" Well, if you're gearing up for exams like JEE Mains, you've probably encountered some pretty intimidating trigonometric equations. Today, we're going to tackle a beast of a problem: finding the values of x in the specific interval [0, 2π] for which the absolute difference between two gnarly square root expressions, specifically ∣√(2sin²x + 18cos²x) - √(2cos²x + 18sin²x)∣, equals 1. Don't sweat it! This isn't just about getting the right answer; it's about understanding the strategy, sharpening your algebraic skills, and building your confidence to tackle similar challenges. We'll break down every single step, making it super clear and easy to follow. So, grab your notebooks, let's dive into this fascinating problem and master complex trigonometric equations together! This kind of problem often appears in competitive exams because it tests multiple concepts: trigonometric identities, algebraic manipulation, solving equations, and understanding domains. Let's get started and turn this seemingly daunting problem into a straightforward victory!

Understanding the Core Problem: Deciphering the Equation

Alright, let's kick things off by really understanding the beast we're up against. Our mission, should we choose to accept it, is to find all values of x within the specific range of [0, 2π] that satisfy the equation ∣√(2sin²x + 18cos²x) - √(2cos²x + 18sin²x)∣ = 1. This equation looks pretty intimidating at first glance, doesn't it? But like any complex problem, it becomes manageable once we break it down into smaller, digestible chunks. The main keywords here are trigonometric equation, absolute value, and square roots. These elements often combine to form tricky questions in exams like JEE Mains.

First off, let's look at the expressions inside the square roots. We have 2sin²x + 18cos²x and 2cos²x + 18sin²x. Notice a pattern here? They're almost mirror images of each other! This symmetry is often a huge hint in math problems, suggesting that there might be some clever simplification waiting to happen. The presence of sin²x and cos²x immediately brings to mind the fundamental trigonometric identity: sin²x + cos²x = 1. This identity is our best friend in almost all trigonometric simplifications, allowing us to convert between sine and cosine squares seamlessly. For instance, we can rewrite cos²x as 1 - sin²x, or sin²x as 1 - cos²x. This flexibility is crucial for unifying our expressions.

Next, we have the absolute value sign: |...| = 1. This means that the expression inside the absolute value can be either +1 or -1. So, we're actually dealing with two potential equations:

  1. √(2sin²x + 18cos²x) - √(2cos²x + 18sin²x) = 1
  2. √(2sin²x + 18cos²x) - √(2cos²x + 18sin²x) = -1

It's super important to consider both cases. For many students, overlooking the negative case of an absolute value equation is a common pitfall. Always remember that |A|=B implies A=B or A=-B.

Finally, the domain or interval for x is [0, 2π]. This means we're looking for solutions within a single full cycle of the trigonometric functions. This constraint is vital because trigonometric functions are periodic, meaning they repeat their values infinitely. Without a specified interval, we'd have an infinite number of solutions. This interval helps us narrow down our search to a specific set of answers. So, once we find our general solutions, we'll need to carefully pick out only those that fall between 0 and 2π (inclusive). Keep in mind that angles like 0, π/2, π, 3π/2, 2π are common boundary points and often critical in these types of problems. Don't forget to include 2π if your general solution allows for it, as the interval is closed. This initial dissection is paramount; it sets the stage for all the algebraic wizardry we're about to perform!

The Algebraic Maneuver: Simplifying the Root Expressions

Okay, guys, now that we've deciphered the core problem, it's time to roll up our sleeves and get into the nitty-gritty of algebraic simplification. This is where we'll turn those complex square root terms into something much more manageable. The goal here is to combine like terms and use our trigonometric identities to reduce the complexity. Our expressions are √(2sin²x + 18cos²x) and √(2cos²x + 18sin²x). Let's tackle each one individually first, then see how they interact. This process is key to unlocking the problem and is a common technique in JEE Mains level questions.

Consider the first term: 2sin²x + 18cos²x. We know that cos²x = 1 - sin²x. Let's substitute that in: 2sin²x + 18(1 - sin²x) 2sin²x + 18 - 18sin²x Combine the sin²x terms: 18 - 16sin²x

Now, let's do the same for the second term: 2cos²x + 18sin²x. Using sin²x = 1 - cos²x: 2cos²x + 18(1 - cos²x) 2cos²x + 18 - 18cos²x Combine the cos²x terms: 18 - 16cos²x

See that? Immediately, the terms look much cleaner! So, our original equation transforms into: ∣√(18 - 16sin²x) - √(18 - 16cos²x)∣ = 1

This is a significant step forward! We've successfully used the identity sin²x + cos²x = 1 to simplify the expressions inside the square roots. Now, notice the symmetry again: 18 - 16sin²x and 18 - 16cos²x. This is where things get really interesting and where the true power of trigonometric identities comes into play. We can relate sin²x and cos²x to cos(2x). Recall the double angle identity: cos(2x) = cos²x - sin²x. We also have cos(2x) = 2cos²x - 1 and cos(2x) = 1 - 2sin²x.

From cos(2x) = 1 - 2sin²x, we can write 2sin²x = 1 - cos(2x), which means sin²x = (1 - cos(2x))/2. From cos(2x) = 2cos²x - 1, we can write 2cos²x = 1 + cos(2x), which means cos²x = (1 + cos(2x))/2.

Let's substitute these into our simplified square root terms. This is a crucial algebraic maneuver that will streamline the rest of the problem.

For 18 - 16sin²x: 18 - 16 * ((1 - cos(2x))/2) 18 - 8(1 - cos(2x)) 18 - 8 + 8cos(2x) 10 + 8cos(2x)

For 18 - 16cos²x: 18 - 16 * ((1 + cos(2x))/2) 18 - 8(1 + cos(2x)) 18 - 8 - 8cos(2x) 10 - 8cos(2x)

Voila! Our equation now looks like this: ∣√(10 + 8cos(2x)) - √(10 - 8cos(2x))∣ = 1

Isn't that much cleaner? We've managed to convert the entire problem from sin²x and cos²x terms into a single trigonometric function: cos(2x). This is a monumental simplification and demonstrates the power of knowing your trig identities cold. This form is often encountered in advanced algebra and trigonometry and sets us up for the next steps where we'll handle the square roots and the absolute value. Keep in mind that for the square roots to be defined, the expressions inside must be non-negative. 10 + 8cos(2x) and 10 - 8cos(2x) must both be ≥ 0. Since cos(2x) ranges from -1 to 1, 8cos(2x) ranges from -8 to 8. So, 10 + 8cos(2x) will be between 2 and 18, and 10 - 8cos(2x) will also be between 2 and 18. Both are always positive, so we don't need to worry about undefined square roots here, which is a relief!

Step-by-Step Simplification: Conquering the Square Roots

Alright, team, we've successfully transformed our intimidating initial problem into a much more approachable form: ∣√(10 + 8cos(2x)) - √(10 - 8cos(2x))∣ = 1. Now, it's time to conquer those square roots and the absolute value. This part requires careful algebraic manipulation and involves squaring both sides of an equation – sometimes even twice! This is a classic technique for eliminating square roots, but it also introduces the possibility of extraneous solutions, so a final verification step will be crucial later on. Pay close attention to each step, as a small error can derail the entire solution, a common challenge in JEE Mains math.

As we discussed earlier, the absolute value |A| = 1 means we have two cases to consider: Case 1: √(10 + 8cos(2x)) - √(10 - 8cos(2x)) = 1 Case 2: √(10 + 8cos(2x)) - √(10 - 8cos(2x)) = -1

Let's start with Case 1: √(10 + 8cos(2x)) - √(10 - 8cos(2x)) = 1. To get rid of one square root, let's isolate one of them and then square both sides. It's usually easier to move the negative term to the other side: √(10 + 8cos(2x)) = 1 + √(10 - 8cos(2x))

Now, square both sides: (√(10 + 8cos(2x)))² = (1 + √(10 - 8cos(2x)))² 10 + 8cos(2x) = 1² + 2 * 1 * √(10 - 8cos(2x)) + (√(10 - 8cos(2x)))² 10 + 8cos(2x) = 1 + 2√(10 - 8cos(2x)) + 10 - 8cos(2x)

Let's simplify and isolate the remaining square root term: 10 + 8cos(2x) = 11 - 8cos(2x) + 2√(10 - 8cos(2x)) Move all non-square root terms to the left side: 10 + 8cos(2x) - 11 + 8cos(2x) = 2√(10 - 8cos(2x)) 16cos(2x) - 1 = 2√(10 - 8cos(2x))

Now, we have a single square root isolated. Time to square both sides again to eliminate it! (16cos(2x) - 1)² = (2√(10 - 8cos(2x)))²

  • (16cos(2x))² - 2 * 16cos(2x) * 1 + 1² = 4 * (10 - 8cos(2x))* 256cos²(2x) - 32cos(2x) + 1 = 40 - 32cos(2x)

Oh, check that out! The -32cos(2x) terms cancel out on both sides. This is often a sign that we're on the right track! 256cos²(2x) + 1 = 40 256cos²(2x) = 39 cos²(2x) = 39/256

This gives us: cos(2x) = ±√(39/256) cos(2x) = ±√39 / 16

Now, let's look at Case 2: √(10 + 8cos(2x)) - √(10 - 8cos(2x)) = -1. Again, isolate one square root. Let's make the term √(10 + 8cos(2x)) + 1 = √(10 - 8cos(2x)). Or, more symmetrically, let's multiply by -1 and rearrange: √(10 - 8cos(2x)) - √(10 + 8cos(2x)) = 1 Notice that this is essentially the same form as Case 1, but with the terms inside the roots swapped. This means if we let Y = -cos(2x), the equation would be √(10 + 8Y) - √(10 - 8Y) = 1. So, the solution for cos(2x) in this case will be the negative of the solution from Case 1.

Let's prove it by following the same steps for Case 2: √(10 + 8cos(2x)) = -1 + √(10 - 8cos(2x)) Square both sides: 10 + 8cos(2x) = 1 - 2√(10 - 8cos(2x)) + 10 - 8cos(2x) 10 + 8cos(2x) = 11 - 8cos(2x) - 2√(10 - 8cos(2x)) 16cos(2x) - 1 = -2√(10 - 8cos(2x))

Now, square both sides again: (16cos(2x) - 1)² = (-2√(10 - 8cos(2x)))² 256cos²(2x) - 32cos(2x) + 1 = 4 * (10 - 8cos(2x)) 256cos²(2x) - 32cos(2x) + 1 = 40 - 32cos(2x) 256cos²(2x) + 1 = 40 256cos²(2x) = 39 cos²(2x) = 39/256 cos(2x) = ±√39 / 16

Phew! Both cases lead to the same result for cos²(2x). This is awesome because it simplifies our job. However, we must remember that when we squared the equation 16cos(2x) - 1 = 2√(10 - 8cos(2x)), we introduced the possibility of extraneous solutions. For Case 1, we needed 16cos(2x) - 1 ≥ 0, which means cos(2x) ≥ 1/16. For Case 2, we needed 16cos(2x) - 1 ≤ 0, which means cos(2x) ≤ 1/16.

Let's check our result cos(2x) = ±√39 / 16. √39 is approximately √36 = 6 and √49 = 7, so roughly 6.24. So, cos(2x) ≈ ±6.24 / 16 ≈ ±0.39. Since 1/16 = 0.0625, both +0.39 and -0.39 satisfy these conditions: 0.39 ≥ 0.0625 (for Case 1) -0.39 ≤ 0.0625 (for Case 2)

This means we need to consider both cos(2x) = +√39 / 16 (from Case 1) and cos(2x) = -√39 / 16 (from Case 2). So, our final target equation is cos(2x) = √39 / 16 from Case 1, and cos(2x) = -√39 / 16 from Case 2. Both solutions for cos(2x) are valid according to their respective intermediate conditions before the second squaring. This is a subtle but critical verification step often missed, leading to incorrect answers in competitive exams.

Solving the Trigonometric Equation: Finding 2x Values

Alright, we've navigated the tricky algebraic terrain and boiled down our original problem to a much simpler trigonometric equation: cos(2x) = ±√39 / 16. This is fantastic! Now, our task is to find the values of 2x that satisfy this equation, and then, from those, derive the values of x within our given interval [0, 2π]. This stage involves a deep understanding of the properties of the cosine function and how to find its general solutions, which is a fundamental skill for JEE Mains and any other advanced math exam.

Let's denote α = arccos(√39 / 16). Since √39 / 16 is a positive value less than 1 (approximately 0.39), α will be an acute angle in the first quadrant, i.e., 0 < α < π/2. Remember, the general solution for cos(θ) = cos(α) is θ = 2nπ ± α, where n is an integer.

So, we have two sub-equations from cos(2x) = ±√39 / 16:

Sub-case A: cos(2x) = √39 / 16 Here, 2x = 2nπ ± α, where α = arccos(√39 / 16).

Sub-case B: cos(2x) = -√39 / 16 Let β = arccos(-√39 / 16). Since (-√39 / 16) is a negative value, β will be an angle in the second quadrant, i.e., π/2 < β < π. Alternatively, we know that cos(π - α) = -cos(α). So, if cos(α) = √39 / 16, then cos(π - α) = -√39 / 16. Therefore, for cos(2x) = -√39 / 16, we can write 2x = 2nπ ± (π - α). It’s often easier to just use the general solution for cos(θ) = k which covers all bases.

Let's list the values of 2x for cos(2x) = √39 / 16: We need to find x in [0, 2π]. This means 2x will be in the interval [0, 4π]. For cos(2x) = √39 / 16 (a positive value), 2x can be in the first or fourth quadrants. The principal value (the arccos value) is α. Possible values for 2x in [0, 4π] are:

  1. 2x = α (First quadrant)
  2. 2x = 2π - α (Fourth quadrant, one full rotation)
  3. 2x = 2π + α (First quadrant, after one full rotation)
  4. 2x = 4π - α (Fourth quadrant, after one full rotation)

These are the four values of 2x for cos(2x) = √39 / 16 within the [0, 4π] range. Remember, we derived this cos(2x) = √39 / 16 from Case 1: √(10 + 8cos(2x)) - √(10 - 8cos(2x)) = 1. This case required cos(2x) ≥ 1/16. Since √39 / 16 ≈ 0.39 is indeed greater than 1/16 ≈ 0.0625, all these solutions for cos(2x) are valid for Case 1.

Now, let's list the values of 2x for cos(2x) = -√39 / 16: For cos(2x) = -√39 / 16 (a negative value), 2x can be in the second or third quadrants. The principal value for this is π - α. Possible values for 2x in [0, 4π] are:

  1. 2x = π - α (Second quadrant)
  2. 2x = π + α (Third quadrant)
  3. 2x = 3π - α (Second quadrant, after one full rotation: 2π + (π - α) = 3π - α)
  4. 2x = 3π + α (Third quadrant, after one full rotation: 2π + (π + α) = 3π + α)

These are the four values of 2x for cos(2x) = -√39 / 16 within the [0, 4π] range. Crucially, we derived this cos(2x) = -√39 / 16 from Case 2: √(10 + 8cos(2x)) - √(10 - 8cos(2x)) = -1. This case required cos(2x) ≤ 1/16. Since -√39 / 16 ≈ -0.39 is indeed less than 1/16 ≈ 0.0625, all these solutions for cos(2x) are valid for Case 2.

In total, we have 8 distinct values for 2x in the interval [0, 4π]. It's super important to be systematic and list them out clearly. Missing a solution, or including an extraneous one, is a common error in JEE Mains calculations. Always double-check your interval boundaries and quadrant logic. This methodical approach ensures you don't miss any valid solutions, and it's a testament to good problem-solving practices.

Finding Solutions within [0, 2π] and Final Verification

We're almost there, guys! We've meticulously worked through the algebra and identified all possible values for 2x within the interval [0, 4π]. Now, the final, crucial step is to convert these 2x values into x values and ensure they fall within our original specified domain, [0, 2π]. This stage also involves a critical verification because squaring operations can sometimes introduce extraneous solutions. In JEE Mains, skipping this verification can lead to wrong answers, so let's be super careful here!

Let's recap our values for 2x: From cos(2x) = √39 / 16 (where α = arccos(√39 / 16)):

  1. 2x = α
  2. 2x = 2π - α
  3. 2x = 2π + α
  4. 2x = 4π - α

From cos(2x) = -√39 / 16 (where we use π - α as the principal value for this negative cosine): 5. 2x = π - α 6. 2x = π + α 7. 2x = 3π - α 8. 2x = 3π + α

Now, let's divide each of these by 2 to get the values for x. Remember, we are looking for x ∈ [0, 2π]. All these 2x values were found within [0, 4π], so dividing them by 2 will naturally place them within [0, 2π].

So, the potential solutions for x are:

  1. x₁ = α/2
  2. x₂ = π - α/2
  3. x₃ = π + α/2
  4. x₄ = 2π - α/2
  5. x₅ = (π - α)/2 = π/2 - α/2
  6. x₆ = (π + α)/2 = π/2 + α/2
  7. x₇ = (3π - α)/2 = 3π/2 - α/2
  8. x₈ = (3π + α)/2 = 3π/2 + α/2

Let's quickly check the order and distinctness. Since 0 < α < π/2, then 0 < α/2 < π/4. This means all these values are distinct and within the [0, 2π] range. For example: x₁ = α/2 is in (0, π/4). x₂ = π - α/2 is in (3π/4, π). x₃ = π + α/2 is in (π, 5π/4). x₄ = 2π - α/2 is in (7π/4, 2π). x₅ = π/2 - α/2 is in (π/4, π/2). x₆ = π/2 + α/2 is in (π/2, 3π/4). x₇ = 3π/2 - α/2 is in (5π/4, 3π/2). x₈ = 3π/2 + α/2 is in (3π/2, 7π/4).

All eight of these values are distinct and lie within [0, 2π].

Now comes the absolutely critical verification step. Remember when we squared both sides? For the solutions derived from cos(2x) = √39 / 16 (i.e., x₁, x₂, x₃, x₄), they came from Case 1 where √(10 + 8cos(2x)) - √(10 - 8cos(2x)) = 1. Before the second squaring, we had the condition 16cos(2x) - 1 ≥ 0, which simplified to cos(2x) ≥ 1/16. Since √39 / 16 ≈ 0.39 is indeed greater than 1/16 ≈ 0.0625, all these four solutions (x₁, x₂, x₃, x₄) are valid because they satisfy the condition cos(2x) = √39/16 > 0 for Case 1.

For the solutions derived from cos(2x) = -√39 / 16 (i.e., x₅, x₆, x₇, x₈), they came from Case 2 where √(10 + 8cos(2x)) - √(10 - 8cos(2x)) = -1. Before the second squaring, we had the condition 16cos(2x) - 1 ≤ 0, which simplified to cos(2x) ≤ 1/16. Since -√39 / 16 ≈ -0.39 is indeed less than 1/16 ≈ 0.0625, all these four solutions (x₅, x₆, x₇, x₈) are valid because they satisfy the condition cos(2x) = -√39/16 < 0 for Case 2.

Since all eight values of x satisfy the conditions that were imposed during the squaring process (which ensured the square roots were positive when we isolated them), all eight solutions are indeed valid solutions to the original equation. Therefore, the number of values of x in the interval [0, 2π] for which the given equation holds true is 8.

This comprehensive verification is what distinguishes a thorough solution from a partial one. In JEE Mains, every detail matters. Understanding why each step is taken, especially during algebraic manipulations like squaring, helps prevent errors and ensures you arrive at the correct final answer. This methodical approach, from problem decomposition to final verification, is a powerful tool for solving complex mathematical problems.

Conclusion

Phew! We made it, guys! We successfully navigated through a challenging trigonometric equation that often appears in competitive exams like JEE Mains. Starting with that intimidating expression, ∣√(2sin²x + 18cos²x) - √(2cos²x + 18sin²x)∣ = 1, we systematically broke it down. We first simplified the terms inside the square roots using the fundamental identity sin²x + cos²x = 1, and then leveraged the double-angle formulas to convert everything into expressions involving cos(2x). This was a critical simplification that allowed us to move forward.

Our journey involved tackling the absolute value by splitting the problem into two distinct cases. We then meticulously applied algebraic techniques to eliminate the square roots by squaring both sides of the equations, not once, but twice! Remember how important it was to isolate the square root terms before squaring to avoid unnecessary complexity? This careful approach led us to the simplified trigonometric equation: cos(2x) = ±√39 / 16.

Finally, we delved into the general solutions of trigonometric equations to find all possible values for 2x within the relevant interval [0, 4π], and then scaled them down to find our x values within [0, 2π]. The final verification step was paramount, where we checked the conditions imposed by our squaring operations to ensure we didn't introduce any extraneous solutions. It's this level of attention to detail that helps secure those extra marks in tough exams.

Ultimately, we found that there are 8 distinct values of x in the interval [0, 2π] that satisfy the given equation. This problem was a fantastic workout for your trigonometric identities, algebraic manipulation skills, and equation-solving prowess. The key takeaways here are to always:

  1. Break down complex problems into smaller, manageable parts.
  2. Utilize fundamental identities to simplify expressions.
  3. Be methodical when dealing with absolute values and square roots.
  4. Always verify your solutions, especially after squaring operations.

Keep practicing these types of problems, and you'll find yourself conquering even the most formidable JEE Mains math challenges with confidence. You've got this! Stay curious, keep learning, and don't hesitate to tackle the next big math adventure!