Problem 14 - 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 the interval \([0,3 \pi]\) satisfying the equation \(2 \sin ^{2} x+5 \sin x-3=0\) is (a) 6 (b) 1 (c) 2 (d) 4

Step-by-Step Solution

Answer

(c) 2

Step 1: Identify the type of the equation

The given equation \(2sin^{2}x + 5sinx - 3 = 0\) is a quadratic equation in sine function.

Step 2: Solve the quadratic sine equation

Rewrite the equation in the format of a quadratic equation: \(2(sin(x))^2 + 5sin(x) - 3 = 0\). Solve the quadratic equation as we normally do. The solutions for sin(x) would then be given by \(sin(x) = [-5 ± sqrt((5^2) - 4*2*(-3))]/(2*2)\). Simplifying, we get \(sin(x) = -1/2, 3\).

Step 3: Identify the values of x for the solutions

We know that \(sin(x) = -1/2\) for \(x = 7π/6, 11π/6, 19π/6, 23π/6, ...\). In the range of [0,3π], only \(x = 7π/6, 11π/6\) satisfies. Similarly, for \(sin(x) = 3\), there are no real solutions as sine values lie in the range [-1,1]. Hence there are no solutions satisfying this condition.

Step 4: Count the number of solutions

As per our calculations, there are only two solutions to the equation in the given interval. Thus, the number of values of x is 2.

Key Concepts

Quadratic Equations

A quadratic equation is an equation of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). This is a standard form taught in algebra, where the highest power of the unknown is 2.
In trigonometry, similar quadratic formats can appear using trigonometric functions, such as sine or cosine, rather than the variable \(x\). In our exercise, the equation \(2\sin^2x + 5\sin x - 3 = 0\) is a quadratic in terms of \(\sin x\).
To solve such an equation, you would use similar methods as solving standard quadratic equations. This typically involves:

• Factoring the quadratic, if possible.
• Using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

This equation needs to be solved for sine, so we can find specific angles \(x\). The quadratic formula is often the fastest and easiest option if factoring is not straightforward.

Trigonometric Functions

Trigonometric functions are functions like sine, cosine, and tangent, which are fundamental in the study of triangles, particularly right-angled triangles. These functions help relate the angles of a triangle to the lengths of its sides.
In our exercise, we focus on the sine function, \( \sin x \), which is a periodic function ranging from -1 to 1. This means any real solution for \(\sin x\) must fall within this interval.
Understanding the properties of sine is crucial:

• It is periodic with a period of \(2\pi\). This means \( \sin(x + 2\pi) = \sin x \).
• Its graph oscillates smoothly between -1 and 1.

When we say \(\sin x = -1/2\), it implies specific values for \(x\) that can be found using known angles, typically within given trigonometric intervals. Knowing the sine values for common angles can help quickly identify solutions.

Interval Solutions

Interval solutions require us to find all solutions within a specific range. Here, the interval given is \([0, 3\pi]\), meaning we need to find all values of \(x\) within this range that satisfy our equation.
Working with intervals stretches our use of periodic properties of trigonometric functions:

• If \(x = 7\pi/6\) and \(x = 11\pi/6\) are solutions for \(\sin x = -1/2\), then due to the periodic nature of sine, we need to cover additional cycles up to \(3\pi\).
• Check if solutions exceed the upper range or are less than the lower range of the interval.

From our step-by-step solution, these are the only valid solutions within \([0, 3\pi]\), which tells us there are precisely two "usable" solutions in this interval. Counting solutions within a specified interval ensures no solutions are missed or unnecessarily counted beyond the interval's limits.