Problem 2 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The general solution of the equation \(\sin ^{50} x-\cos ^{50} x=1\) is (A) \(2 n \pi+\frac{\pi}{2}\) (B) \(2 n \pi+\frac{\pi}{3}\) (C) \(n \pi+\frac{\pi}{2}\) (D) \(n \pi+\frac{\pi}{3}\)

Step-by-Step Solution

Answer

The correct answer is (A) \(2 n \pi + \frac{\pi}{2}\).

Step 1: Understand the Equation

We have the equation \(\sin^{50} x - \cos^{50} x = 1\). This means we need to find values of \(x\) for which this equality holds.

Step 2: Analyze the Range of Sine and Cosine

Since both \(\sin(x)\) and \(\cos(x)\) range from -1 to 1, analyze the extremal values to understand that the equation equals 1 if \(\sin^{50}x = 1\) and \(\cos^{50}x = 0\).

Step 3: Solve \(\sin(x) = 1\)

The equation \(\sin(x) = 1\) translates to \(x = \frac{\pi}{2} + 2n\pi\) (where \(n\) is an integer) because this is where the sine function achieves its maximum value.

Step 4: Verify Conditions

At \(x = \frac{\pi}{2} + 2n\pi\), \(\cos(x) = 0\) satisfies \(\cos^{50} x = 0\). Thus, the conditions for the equation \(\sin^{50} x - \cos^{50} x = 1\) are satisfied.

Step 5: Conclusion and Choose The Correct Option

The general solution \(x = \frac{\pi}{2} + 2n\pi\) matches with option (A): \(2n\pi + \frac{\pi}{2}\).

Key Concepts

General Solution

In trigonometry, finding a general solution involves determining a formula that includes all possible solutions for an equation. In the context of the given problem, we are tasked with finding the solution to the equation \( \sin^{50} x - \cos^{50} x = 1 \). To do this, we analyze the conditions under which the equation holds true. Since the sine and cosine functions are periodic, their general solutions involve adding integer multiples of their respective periods.

• The sine function, \( \sin(x) \), reaches its maximum value of 1 at specific points. Hence, solving \( \sin(x) = 1 \) aids in solving our main equation.
• The cosine function needs to become zero to satisfy \( \cos^{50}x = 0 \), which happens at the same points where \( \sin(x) = 1 \).

As a result, we reach the conclusion that the general solution is given by \( x = \frac{\pi}{2} + 2n\pi \). This captures all instances, where \( n \) is any integer, meaning the pattern repeats every full cycle of the sine function, encapsulated by \(2\pi\).

Sine Function

The sine function, represented as \( \sin(x) \), is a fundamental trigonometric function that describes oscillations and waves in mathematics. It is important to remember:

• \( \sin(x) \) varies between -1 and 1.
• Its typical graph spans a complete cycle over the interval from 0 to \( 2 \pi \).
• Maxima occur where \( \sin(x) = 1 \), which plays a crucial role in solving equations like our original one.

In our initial equation \( \sin^{50}x - \cos^{50}x = 1 \), for the equation to hold, \( \sin(x) \) must be 1, meaning \( x \) aligns with points where the sine curve hits its peak. Thus, \( x = \frac{\pi}{2} + 2n\pi \) captures when sine is at maximum, ensuring the equation is satisfied.

Cosine Function

The cosine function, represented as \( \cos(x) \), complements the sine function with the following characteristics:

• Like sine, \( \cos(x) \) varies between -1 and 1.
• The function completes a cycle as \( x \) progresses over an interval from 0 to \( 2\pi \).
• Zero points occur where \( \cos(x) = 0 \), crucial to solving certain equations.

In our scenario of solving \( \sin^{50}x - \cos^{50}x = 1\), it's essential that \( \cos(x) \) equals zero. This coincides with peaks of the sine function, specifically at \( x = \frac{\pi}{2} + 2n\pi \). The cosine function is zero at angles where the sine function is at its maximum, providing a perfect condition for satisfying our given equation, and thus ensuring \( \cos^{50}x \to 0 \).