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

Question

The general solution of the equation \(\sin ^{50} x-\cos ^{30} x=1\) 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 general solution is \(2n\pi + \frac{\pi}{2}\), so the answer is (A).

Step 1: Identify the Range of Trigonometric Functions

Understand that since \(-1 \leq \sin x, \cos x \leq 1\), the maximum value of \(\sin^{50} x\) is \(1\) and the maximum value of \(\cos^{30} x\) is also \(1\).

Step 2: Analyze the Equation

Given the equation is \(\sin^{50} x - \cos^{30} x = 1\). For this equation to hold true, \(\sin^{50} x\) must be equal to \(1\) and \(\cos^{30} x\) must be equal to \(0\) since 1 - 0 is 1.

Step 3: Find Conditions for \(\sin^{50} x = 1\)

For \(\sin^{50} x = 1\), it is necessary that \(\sin x = 1\). This implies that \(x = \frac{\pi}{2} + 2n\pi\) for integer values of \(n\).

Step 4: Find Conditions for \(\cos^{30} x = 0\)

For \(\cos^{30} x = 0\), it is necessary that \(\cos x = 0\). This implies that \(x = \frac{\pi}{2} + n\pi\) for integer values of \(n\).

Step 5: Determine the General Solution

The general solution must satisfy both conditions in Step 3 and Step 4, which coincide for \(x = \frac{\pi}{2} + 2n\pi\). Thus, the solution encompasses only the values where both sine and cosine conditions overlap perfectly.

Key Concepts

General Solution

When solving trigonometric equations, finding the general solution is essential. The general solution represents all possible solutions for a trigonometric equation over its entire range.
To find the general solution, we determine a pattern of values for the variable that satisfies the equation each time within one full cycle of the trigonometric function. In this example, we were given the equation \( \sin ^{50} x - \cos ^{30} x = 1 \).
Analyzing possible solutions, it was necessary to see when both \( \sin x = 1 \) and \( \cos x = 0 \).
The pattern that satisfies both conditions in the equation was determined to be \( x = \frac{\pi}{2} + 2n\pi \) for any integer \( n \). Hence, identifying the right conditions that align both trigonometric functions gives us the general solution.

Sine Function

The sine function, often noted as \( \sin(x) \), is a foundational concept in trigonometry.
It is periodic with a period of \( 2\pi \), meaning \( \sin(x) = \sin(x + 2\pi) \).
This periodic nature is crucial when determining solutions like in our given equation. The definition is based on a right triangle, where \( \sin(x) \) is the ratio of the opposite side to the hypotenuse.

• Maximum value of 1 occurs at \( x = \frac{\pi}{2} + 2n\pi \).
• Graphically, it resembles a wave starting at zero, peaking at \( \pi/2 \), and repeating.

In our solution, we determined that \( \sin x = 1 \) was required for the equation value, which occurs at specific points like \( \frac{\pi}{2} \) plus multiples of \( 2\pi \).
This insight helped us derive parts of the general solution.

Cosine Function

Like the sine function, the cosine function \( \cos(x) \) is also a periodic trigonometric function.
It shares the same period as the sine function, that is, \( 2\pi \). Graphically, it starts at its maximum and forms a wave similar to sine, but shifted.
In terms of triangles, \( \cos(x) \) is the ratio of the adjacent side to the hypotenuse.

• Its maximum value is 1 when \( x = 0 + 2n\pi \).
• Zero occurs at \( x = \frac{\pi}{2} + n\pi \).

For our exercise, setting \( \cos^{30} x = 0 \) implied \( \cos x = 0 \), leading to \( x = \frac{\pi}{2} + n\pi \).
This calculation is vital for matching the sine solution's requirements.

Range of Trigonometric Functions

Understanding the range of trigonometric functions is key to solving equations like the one we've tackled.
The range indicates all possible values the function can achieve.

• For \( \sin x \) and \( \cos x \), the range is \([-1, 1]\).
• This limits what certain transformations of \( \sin \) and \( \cos \) can equate to \( 1 \) or \( 0 \).

In our original problem, only values of \( 1 \) (for \( \sin x \)) and \( 0 \) (for \( \cos x \)) were investigated, as these satisfied the trigonometric equation \( \sin^{50} x \) and \( \cos^{30} x \) necessity.
Understanding these bounds is essential, as it helps realize why other potential values cannot solve the equation. These limitations are why precise equations must be evaluated thoroughly within these ranges.