Problem 15 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(\cos 3 x+\sin \left(2 x-\frac{7 \pi}{6}\right)=-2\) then \(x=\) (A) \(\frac{\pi}{3}(6 k+1)\) (B) \(\frac{\pi}{3}(6 k-1)\) (C) \(\frac{\pi}{3}(2 k+1)\) (D) none of these where \(k \in Z\)

Step-by-Step Solution

Answer

(B) \(x = \frac{\pi}{3}(6k - 1)\)

Step 1: Analyze the Trigonometric Equation

We need to solve the equation \( \cos(3x) + \sin\left(2x - \frac{7\pi}{6}\right) = -2 \). The maximum value that the sum of \( \cos(3x) \) and \( \sin\left(2x - \frac{7\pi}{6}\right) \) can take is \( 2 \), and the minimum value is \( -2 \). Therefore, for the equation to equal \(-2\), each function must equal \(-1\).

Step 2: Solve \(\cos(3x) = -1\)

The cosine function equals \(-1\) at \(3x = \pi + 2m\pi\) for any integer \(m\). Solving for \(x\), we get: \[ x = \frac{\pi}{3}(2m + 1). \]

Step 3: Solve \(\sin\left(2x - \frac{7\pi}{6}\right) = -1\)

The sine function equals \(-1\) at \(2x - \frac{7\pi}{6} = \frac{3\pi}{2} + 2n\pi\), where \(n\) is an integer. Rewriting, we have: \( 2x = \frac{3\pi}{2} + \frac{7\pi}{6} + 2n\pi \). This simplifies to \[ 2x = \frac{16\pi}{6} + 2n\pi = \frac{8\pi}{3} + 2n\pi. \] Therefore, \[ x = \frac{4\pi}{3} + n\pi. \]

Step 4: Equate and Solve for Common Solutions

From Steps 2 and 3, we have two general solutions for \(x\): \( x = \frac{\pi}{3}(2m + 1) \) and \( x = \frac{4\pi}{3} + n\pi \). Equate these to find common solutions: \[ \frac{\pi}{3}(2m + 1) = \frac{4\pi}{3} + n\pi. \] Simplifying gives: \[ 2m + 1 = 4 + 3n. \] Rewriting: \[ 2m = 3n + 3 \] or \[ m = \frac{3n + 3}{2}. \] Letting \( n = 2p + 1 \) (since \(3n\) must be odd), we get \( m = 3p + 2 \). Substituting \( m \) back, \[ x = \frac{\pi}{3}(6p + 5). \]

Step 5: Check Solution Against Multiple Choice Options

Compare \( x = \frac{\pi}{3}(6p + 5) \) to the provided options. Substitute \( k = p \), and it matches the form of option (B): \( x = \frac{\pi}{3}(6k + 5) \), simplifying to (B): \( x = \frac{\pi}{3}(6k - 1) \) after further inspection, realizing option inconsistency.

Key Concepts

Trigonometric Functions

Trigonometric functions are crucial in expressing angles and their relationships. The primary trigonometric functions include sine \( \sin \), cosine \( \cos \), and tangent \( \tan \). \textbf{Sine} of an angle in a right triangle is defined as the ratio of the length of the opposite side to the hypotenuse, whereas \textbf{cosine} is the ratio of the adjacent side to the hypotenuse. \textbf{Tangent} is the ratio between the opposite side and the adjacent side, and can also be expressed as \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \).

In unit circle terms, these functions describe the coordinates of angles on a circle with a radius of one. For example, \( \cos(\theta) \) gives the x-coordinate of the point on the unit circle, while \( \sin(\theta) \) gives the y-coordinate. These functions are periodic, meaning they repeat values in regular intervals. The periodicity helps solve more complex trigonometric equations.

Maximum and Minimum Values of Trigonometric Functions

Understanding the maximum and minimum values of trigonometric functions is essential in solving equations like \( \cos(3x) + \sin\left(2x - \frac{7\pi}{6}\right) = -2 \). Both the sine and cosine functions have a range of values from -1 to 1, resulting in a combined range of -2 to 2 for their sum.

The maximum value of \( \cos(3x) + \sin\left(2x - \frac{7\pi}{6}\right) \) is 2 when each function reaches its peak at 1. Conversely, the minimum value is -2 when both functions are at -1. This property allows us to decide the behavior and solutions of the function. When a problem specifies the sum equals -2, it implies both trigonometric components \( \cos(3x) \) and \( \sin\left(2x - \frac{7\pi}{6}\right) \) must equal -1 simultaneously for the condition to hold true.

Solving Trigonometric Equations

When solving trigonometric equations, one key approach is equating the functions to their known maximum or minimum values to find solutions. In the equation \( \cos(3x) + \sin\left(2x - \frac{7\pi}{6}\right) = -2 \), individual solutions must be found where each component hits its minimum value of -1.

\begin{itemize} \item First, solve \( \cos(3x) = -1 \) which occurs at \( 3x = \pi + 2m\pi \). We find \( x = \frac{\pi}{3}(2m+1) \). \item Next, solve \( \sin\left(2x - \frac{7\pi}{6}\right) = -1 \), occurring at \( 2x - \frac{7\pi}{6} = \frac{3\pi}{2} + 2n\pi \), giving \( x = \frac{4\pi}{3} + n\pi \). \end{itemize}
By equating these two expressions for \( x \), \( \frac{\pi}{3}(2m+1) = \frac{4\pi}{3} + n\pi \), the goal is to find common solutions. Simplifying this relationship leads to feasible values that fulfill both conditions, often dictated by setting parameters for integers such as \( m \) and \( n \). \textbf{Remember}, checking solutions against multiple-choice options ensures arriving at the correct interpretation of numerical expressions, slightly adjusting form-dependent expressions if necessary.