Problem 51 - Comprehensive Trigonometry for IIT JEE Main and Advanced Rejaul Makshud MeGraw Hill

Question

Solve the following equations and tick the correct one. The number of solutions of the equation \(\tan x+\sec x=2 \cos x\) lying in the interval \([0,2 \pi]\) is (a) 0 (b) 1 (c) 2 (d) 3

Step-by-Step Solution

Answer

The number of solutions of the equation \( \tan x+\sec x=2 \cos x \) lying in [0,2\pi] is 3

Step 1: Rewrite the equation

Rewrite the equation from \( \tan x+\sec x=2 \cos x \) to \( \frac{\sin x}{\cos x} + \frac{1}{\cos x} - 2 \cos x = 0 \), as the definitions for the trigonometric functions are \(\tan x = \frac{\sin x}{\cos x}\), \(\sec x = \frac{1}{\cos x} \) and \(\cos x \) remains as it is.

Step 2: Simplify the equation

Combine terms to simplify the equation from \( \frac{\sin x}{\cos x} + \frac{1}{\cos x} - 2 \cos x = 0 \) to \( \frac{\sin x + 1 - 2 \cos^2 x}{\cos x} = 0 \). After factoring the \(\cos x \) out of the equation in the denominator, it is noticed that \(\cos^2 x = 1 - \sin^2 x \) by Pythagorean identity, replace \(\cos^2 x \) with \(1 - \sin^2 x\), obtaining \( \frac{\sin x + 1 - 2 + 2 \sin^2 x}{\cos x} = 0 \) which simplifies further into \( \frac{2\sin^2 x + \sin x - 1}{\cos x} = 0 \).

Step 3: Solve the quadratic equation

The equation \( \frac{2\sin^2 x + \sin x - 1}{\cos x} = 0 \) is a quadratic equation in sine. Solve the equation as a quadratic equation in \(\sin x\). The solutions are \( \sin x = -1 \) and \( \sin x = \frac{1}{2}\). Now locate solutions in the interval given [ 0, 2\pi].

Step 4: Determine the number of solutions

For \(\sin x = -1\), there is 1 solution in [0, 2\pi] which is \(x=\pi\) . For \(\sin x = \frac{1}{2}\), there are 2 solutions in [0, 2\pi] which are \(x = \frac{\pi}{6}, \frac{5\pi}{6}\). So, there are a total of 3 solutions which belong to given interval [0, 2\pi].

Step 5: Tick the correct choice

From according to the number of solutions found (3), the correct option from the multiple choices given is (d) 3.

Key Concepts

Solutions of Trigonometric Equations

Trigonometric equations involve one or more trigonometric functions such as sine, cosine, or tangent. Solving these equations typically requires finding the value of the variable that will satisfy the equation. In our problem, we have the equation \(\tan x + \sec x = 2 \cos x\), which we need to solve within the interval \([0, 2\pi]\). This means that we are looking for all angles \(x\) that fit these criteria.

• The solutions can suggest points on a unit circle whereby the angle \(x\) leads to the same value on both sides of the equation.
• Once the equation is solved, the solutions must be verified to be within the specified interval.

Specific to our equation, we convert tangents and secants to sine and cosine forms, allowing us to work with more fundamental trigonometric identities. Each solution represents an angle where both the trigonometric expressions equalize.

Solving Quadratic Equations in Trigonometric Functions

Solving quadratic equations in trigonometric functions requires utilizing techniques similar to solving standard quadratic equations. The primary difference lies in the nature of the trigonometric functions involved. In our example, after transforming the initial equation, we end up with a quadratic in sine: \(2\sin^2 x + \sin x - 1 = 0\).

• To solve this quadratic equation, we apply the quadratic formula: \(\sin x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 2\), \(b = 1\), and \(c = -1\).
• The solutions derived are \(\sin x = -1\) and \(\sin x = \frac{1}{2}\).

Since \(\sin x\) must be within a real range, each solution needs to checked within \([0, 2\pi]\). For instance, \(\sin x = -1\) corresponds to \(x = \pi\), and \(\sin x = \frac{1}{2}\) corresponds to \(x = \frac{\pi}{6}\) and \(x = \frac{5\pi}{6}\). These calculations identify all possible angles in the specified interval.

Trigonometric Identities and Simplification

Trigonometric identities are formulas that allow us to relate different trigonometric functions to one another. These identities are crucial for simplifying complex trigonometric expressions to make them solvable. In our situation, we converted \(\tan x\) and \(\sec x\) in terms of sine and cosine functions to simplify the equation.

• The identity \(\tan x = \frac{\sin x}{\cos x}\) helps convert tangents which are often easier to handle in sine and cosine terms.
• Similarly, \(\sec x = \frac{1}{\cos x}\) aids in converting secant functions, facilitating simplification.
• Utilizing the Pythagorean identity, \(\cos^2 x = 1 - \sin^2 x\), further helped in reducing the complexity of the original equation.

These identities enable simplification, leading the way to form a solvable quadratic equation. Mastery of these identities is essential for solving a wide range of trigonometric problems efficiently.