Problem 15 - 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 values of \(x\) in \([0,5 \pi]\) satisfying the equation \(3 \sin ^{2} x-7 \sin x+2=0\) is (a) 0 (b) 5 (c) 6 (d) 10

Step-by-Step Solution

Answer

The correct answer is that there are 4 solutions. Hence, no options given in the exercise are correct.

Step 1: Identify the equation type

The equation \(3 \sin^2{x} - 7\sin{x} + 2 = 0\) is a quadratic equation in terms of \(\sin{x}\). It is given in the form \(a\sin^2{x} + b\sin{x} + c = 0\) where \(a = 3\), \(b = -7\), and \(c = 2\).

Step 2: Apply the quadratic formula

To find the roots of this quadratic equation, we apply the quadratic formula \(\sin{x} = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\). Substituting the values \(a = 3\), \(b = -7\), and \(c = 2\), we get \(\sin{x} = \frac{7 \pm \sqrt{(-7)^2-4 \cdot 3 \cdot 2}}{2\cdot 3} = \frac{7 \pm 1}{6}\). Hence, the roots are \(\sin{x} = \frac{4}{3}\), which is not possible as the value of \(\sin{x}\) cannot exceed 1, and \(\sin{x} = 1\), which is a valid solution.

Step 3: Find all solutions for x

The value \(\sin{x} = 1\) corresponds to \(x = \frac{\pi}{2} + 2k\pi\). We need to find all the values of \(x\) in the interval [0, 5\(\pi\)]. For \(k = 0\), there is one solution \(x = \frac{\pi}{2}\). For \(k = 1\), there are two solutions \(x = \frac{\pi}{2} + 2\pi\) and \(x = \frac{5\pi}{2} + 2\pi\). For \(k = 2\), there is one solution \(x = \frac{\pi}{2} + 4\pi\). Therefore, there are a total of 4 solutions in this interval.

Key Concepts

Quadratic Equations

Quadratic equations are fundamental in mathematics and often occur in various contexts, including physics and engineering.
These equations are usually written in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) represents the unknown variable. In the context of trigonometric equations, the variable can be a trigonometric function such as \( \sin{x} \) or \( \cos{x} \).
To solve quadratic equations, one common method is the quadratic formula given by:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This formula allows us to find the roots of the equation, providing solutions for the unknown variable, \( x \). Recognizing a quadratic equation within another context, such as in terms of \( \sin{x} \), is essential to applying this formula correctly and finding all possible solutions.
For example, the equation \(3 \sin^2{x} - 7\sin{x} + 2 = 0\) is quadratic in \( \sin{x} \), and hence the quadratic formula can be applied to find values of \( \sin{x} \) that satisfy the equation.

Trigonometric Functions

Trigonometric functions describe relationships involving angles and lengths in right-angled triangles.
The primary trigonometric functions include sine, cosine, and tangent, each defined in terms of a right-angled triangle.

• \( \sin{x} = \frac{\text{opposite side}}{\text{hypotenuse}} \)
• \( \cos{x} = \frac{\text{adjacent side}}{\text{hypotenuse}} \)
• \( \tan{x} = \frac{\text{opposite side}}{\text{adjacent side}} \)

In trigonometric equations like the one in the original exercise, often constraints apply.
For instance, the sine function always produces values between \(-1\) and \(1\). Therefore, any value resulting from solving a trigonometric equation that doesn't lie within this interval cannot be a valid solution.
Knowing the periodic nature of these functions is also helpful. For \( \sin{x} \), the function repeats its values every \(2\pi\) units, which is crucial when determining all possible solutions within a specified interval. This periodic behavior helps in accounting for additional solutions derived from the base angle.

Interval Notation

Interval notation is a concise way of describing a range of numbers along a number line.
It is used extensively in mathematics to specify the domain or solutions of a function or equation.
For an interval from point \(a\) to point \(b\), we might write:

• \([a, b]\) if both endpoints \(a\) and \(b\) are included (closed interval)
• \((a, b)\) if \(a\) and \(b\) are not included (open interval)
• \([a, b)\) or \((a, b]\) if one end is included and the other is not (half-open intervals)

The interval \([0, 5\pi]\) from the exercise specifies that values are considered starting from 0 up to and including \(5\pi\).
This notation is essential when determining which solutions of a trigonometric equation fall within a given range.
It ensures that none of the potential solutions are outside the boundaries defined by the problem. Knowing how to interpret and apply interval notation is key for verifying the solutions to equations, especially when they include infinite or unbounded ranges in trigonometric functions.