Problem 43 - A Complete Resource Book in Mathematics for JEE Main

Question

Solution of the equation \(\sin 6 x+\cos 4 x+2=0 ; 0

Step-by-Step Solution

Answer

The solution is (B) \(x = \frac{\pi}{4}\).

Step 1: Rearrange the Equation

The given equation is \( \sin(6x) + \cos(4x) + 2 = 0 \). Let's rearrange it as \( \sin(6x) + \cos(4x) = -2 \).

Step 2: Analyze the Range of the Trigonometric Functions

The functions \( \sin \theta \) and \( \cos \theta \) both range from \(-1\) to \(1\). Therefore, the sum \( \sin(6x) + \cos(4x) \) ranges from \(-2\) to \(2\). This implies the sum can indeed equal \(-2\).

Step 3: Find Conditions for the Sum to Equal -2

For \( \sin(6x) + \cos(4x) = -2 \), both \( \sin(6x) \) and \( \cos(4x) \) must be \(-1\) because they can independently range from \(-1\) to \(1\). Thus, \( \sin(6x) = -1 \) and \( \cos(4x) = -1 \).

Step 4: Solve \( \sin(6x) = -1 \)

The equation \( \sin(\theta) = -1 \) has solutions when \( \theta = \frac{3\pi}{2} + 2n\pi \) for integer \(n\). Therefore, for \( \sin(6x) = -1 \), \(6x = \frac{3\pi}{2} + 2n\pi\). Solving for \(x\), we get \(x = \frac{3\pi}{12} + \frac{2n\pi}{6} = \frac{\pi}{4} + \frac{n\pi}{3}\).

Step 5: Solve \( \cos(4x) = -1 \)

The equation \( \cos(\phi) = -1 \) has solutions when \( \phi = \pi + 2m\pi \) for integer \(m\). Thus, for \( \cos(4x) = -1 \), we solve \(4x = \pi + 2m\pi\). Solving for \(x\), we find \(x = \frac{\pi}{4} + \frac{m\pi}{2}\).

Step 6: Find the Common Solution

The solutions for both equations must be identical. We have \(x = \frac{\pi}{4} + \frac{n\pi}{3}\) and \(x = \frac{\pi}{4} + \frac{m\pi}{2}\). Find common values by equating: \(\frac{\pi}{4} + \frac{n\pi}{3} = \frac{\pi}{4} + \frac{m\pi}{2}\). This results in \(\frac{n\pi}{3} = \frac{m\pi}{2}\). Solve for integer values of \(n\) and \(m\) to satisfy \(0 < x < 2\pi\).

Step 7: Solution Verification

We deduced that \(x = \frac{\pi}{4} + n \cdot \frac{\pi}{3} = \frac{\pi}{4} + 0 = \frac{\pi}{4}\); this matches choice (B). Verify within the bounds \(0 < x < 2\pi\), where this is true since \(\frac{\pi}{4} < 2\pi\).

Key Concepts

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are fundamental in mathematics and are periodic by nature. Each of these functions take an angle, often measured in radians or degrees, as input and output a value between -1 and 1.
- **Sine Function**: Represented as \( \sin(\theta) \), it calculates the y-coordinate of a point on a unit circle.- **Cosine Function**: Represented as \( \cos(\theta) \), it calculates the x-coordinate of a point on a unit circle.Both \(\sin(\theta)\) and \(\cos(\theta)\) are periodic with a period of \(2\pi\), meaning \( \sin(\theta + 2\pi) = \sin(\theta) \) and \( \cos(\theta + 2\pi) = \cos(\theta) \). By understanding these characteristics, you can easily predict their values at specific intervals and solve related equations.

Solution of Equations

Solving trigonometric equations involves finding all the values of the variable that satisfy the equation within a given range. This process often requires

• Identifying corresponding angles that satisfy basic trigonometric identities, such as \( \sin\theta = -1 \) or \( \cos\phi = -1 \).
• Utilizing general solutions for trigonometric functions.

For instance, \( \sin(\theta) = -1 \) indicates an angle \( \theta = \frac{3\pi}{2} + 2n\pi \) for integer \(n\). Similarly, when solving for \( \cos(\theta) = -1 \), you recognize \( \theta = \pi + 2m\pi \) for integer \(m\).
Combining these, you find intersections within defined limits to identify valid solutions. For example, the solutions must match the condition \( 0 < x < 2\pi \) to suit a typical trigonometric equation problem.

Range of Functions

The range of trigonometric functions is crucial since it determines the possible output values for specific inputs. For both sine and cosine functions:

• The range is \(-1\)\ to \(+1\)\.
• This means that any linear combination must also respect these bounds.

To solve an equation like \( \sin(6x) + \cos(4x) = -2 \), recognizing the combined range, which extends from \(-2\) to \(+2\) due to the bounds of each function component, allows you to ensure such a solution is possible. Understanding how these function limits work helps to investigate conditions where extreme values are achieved, like when each function equals \( -1 \) in tandem.
By grasping these ranges, students can focus on plausible solutions under specific constraints and verify the potential for such results based on function limitations.