Problem 25 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

Solve: \(\sin 2 x+\sin x+\cos 2 x+\cos x+1=0\)

Step-by-Step Solution

Answer

The solutions to the equation are \(x = 2\pi/3 + 2k\pi\), \(x = 4\pi/3 + 2k\pi\), and \(x = \pi/2 + 2k\pi\), where \(k\) is any integer.

Step 1: Apply double angle formulas

Let's use the double angle formulas: \(\sin2x = 2\sin x\cos x\) and \(\cos2x = \cos^2x - \sin^2x\). Applying these identities to the given equation, we get: \(2\sin x\cos x + \sin x + \cos^2x - \sin^2x + \cos x + 1 = 0\)

Step 2: Combine like terms

Next, rearrange and collect the terms to make it easy to handle. We can combine \(\sin x\) and \(2\sin x\cos x\) to get us \( \sin x(2\cos x + 1)\), and similarly for the cosine term and the constants, so we get: \(\sin x(2\cos x + 1) + \cos x(1 - \sin x) + 1 = 0\)

Step 3: Simplify the equation

To solve the trigonometric equation, we set each of the terms inside brackets to zero, and solve for \(x\). \(2\cos x + 1 = 0 \Rightarrow \cos x = -1/2\), \(1 - \sin x = 0 \Rightarrow \sin x = 1\), and lastly checking if the term \(\cos x(1 - \sin x) + 1\) can be 0.

Step 4: Find solutions in the unit circle

The solutions for \(\cos x = -1/2\) are \(x = 2\pi/3 + 2k\pi\) and \(x = 4\pi/3 + 2k\pi\), where \(k\) is an integer. The solution for \(\sin x = 1\) is \(x = \pi/2 + 2k\pi\). Checking the term \(\cos x(1 - \sin x) + 1\), we find that there are no real solutions.

Key Concepts

Double Angle Identities

Double angle identities are fundamental in solving trigonometric equations. These identities are used to transform trigonometric functions involving double angles, like \(2x\), into expressions that involve single angles, such as \(x\). This transformation makes the equation easier to manage and solve.
A couple of the most commonly used double angle identities include:

• \(\sin 2x = 2\sin x\cos x\)
• \(\cos 2x = \cos^2 x - \sin^2 x\)

In the given exercise, we replaced \(\sin 2x\) with \(2\sin x \cos x\) and \(\cos 2x\) with \(\cos^2x - \sin^2x\) to simplify the original equation. This substitution helps in identifying and combining like terms, moving us a step closer to the solution.

Unit Circle

The unit circle is a crucial tool when working with trigonometric equations, particularly when finding solutions for angles expressed in radians. Imagine a circle with a radius of one centered at the origin of a coordinate plane. Each point on the circle corresponds to specific values of sine and cosine.
On the unit circle:

• \(\cos x\) corresponds to the x-coordinate of a point.
• \(\sin x\) corresponds to the y-coordinate of a point.

For the equation \(\cos x = -1/2\), we find two points on the unit circle that correlate: \(x = 2\pi/3\) and \(x = 4\pi/3\), and for \(\sin x = 1\), the solution is \(x = \pi/2\). Factoring in periodicity, we add \(2k\pi\) to denote additional solutions obtained by rotating around the circle, where \(k\) is an integer.

Trigonometric Solutions

Trigonometric solutions in equations often involve finding angle values that satisfy the given equation. By reducing the equation using identities and simplifying terms, we find the basic solutions which correspond to specific points on the unit circle.
In our exercise, we initially found two distinct conditions: \(2\cos x + 1 = 0\) and \(1 - \sin x = 0\), offering solutions based on known trigonometric values:

• For \(\cos x = -1/2\), the solutions are \(x = 2\pi/3 + 2k\pi\) and \(x = 4\pi/3 + 2k\pi\).
• For \(\sin x = 1\), the solution is \(x = \pi/2 + 2k\pi\).

We checked whether the other conditions yield solutions or contradictions, a critical step to ensure the correct and complete solving of the trigonometric equation.