Problem 27 - Calculus with Concepts in Calculus

Question

Solve the equation \(2 \sin ^{2} x+\sin x-1=0\).

Step-by-Step Solution

Answer

The solutions are \(x = \frac{\pi}{6} + 2k\pi\), \(x = \frac{5\pi}{6} + 2k\pi\), and \(x = \frac{3\pi}{2} + 2k\pi\), where \(k \in \mathbb{Z}\).

Step 1: Recognize the Equation Type

We start with the trigonometric equation \(2 \sin^2 x + \sin x - 1 = 0\). This is a quadratic equation in terms of \(\sin x\).

Step 2: Set a Substitution

To solve the equation, let \(u = \sin x\). The equation becomes \(2u^2 + u - 1 = 0\).

Step 3: Apply the Quadratic Formula

The quadratic equation \(au^2 + bu + c = 0\) can be solved using the quadratic formula: \[u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]. Here, \(a = 2\), \(b = 1\), and \(c = -1\).

Step 4: Calculate the Discriminant

First, calculate the discriminant \(b^2 - 4ac\): \(1^2 - 4 \cdot 2 \cdot (-1) = 1 + 8 = 9\).

Step 5: Solve for u

Substitute the values into the quadratic formula: \[u = \frac{-1 \pm \sqrt{9}}{4} = \frac{-1 \pm 3}{4}\]. This gives \(u = \frac{2}{4} = \frac{1}{2}\) or \(u = \frac{-4}{4} = -1\).

Step 6: Find the Solutions for x

Since \(u = \sin x\), solve \(\sin x = \frac{1}{2}\) and \(\sin x = -1\). For \(\sin x = \frac{1}{2}\), \(x = \frac{\pi}{6} + 2k\pi\) and \(x = \frac{5\pi}{6} + 2k\pi\) where \(k\) is an integer. For \(\sin x = -1\), \(x = \frac{3\pi}{2} + 2k\pi\).

Key Concepts

Quadratic Formula

The quadratic formula is a versatile tool used in algebra to solve equations of the form \( ax^2 + bx + c = 0 \). It allows us to find the values of \( x \) that satisfy the equation. The formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]In the original exercise, we had the equation \( 2\sin^2 x + \sin x - 1 = 0 \) that resembled the form \( ax^2 + bx + c = 0 \) with \( u = \sin x \). By substituting \( u \), we transformed our trigonometric equation into a more familiar quadratic equation, allowing us to apply this formula.

• -b: Represents the additive inverse of the coefficient of the linear term.
• \( \pm \sqrt{b^2 - 4ac} \): Introduces the concept of the discriminant, which determines the nature of the roots.
• 2a: Acts as the divisor for the entire expression, ensuring we are solving for the values of the variable precisely.

Trigonometric Functions

Trigonometric functions, such as sine (\( \sin \)), cosine (\( \cos \)), and tangent (\( \tan \)), are essential in understanding relationships within a right-angled triangle. In our exercise, we are particularly focused on the sine function.The sine function, \( \sin x \), describes the ratio of the opposite side to the hypotenuse in a right-angled triangle. In mathematical terms, it can take values from -1 to 1. For the equation \( \sin x = \frac{1}{2} \), we look for angles where the sine function gives \(\frac{1}{2}\). These angles are \( x = \frac{\pi}{6} \) and \( x = \frac{5\pi}{6} \) in the unit circle.

• \( \sin x = \frac{1}{2} \) occurs at specific standard angles in the first and second quadrant due to sine's periodic nature.
• \( \sin x = -1 \) specifically occurs at \( x = \frac{3\pi}{2} \), illustrating the function's behavior at distinct values.

Discriminant

The discriminant in the quadratic formula, represented as \( b^2 - 4ac \), is crucial for understanding the nature of the solutions to a quadratic equation. It helps in predicting whether the roots are real or complex, and if they are real, whether they are distinct or repeated.

Understanding the Discriminant:

• Positive Discriminant: Indicates two distinct real solutions. This case appeared in our exercise with a discriminant of 9.
• Zero Discriminant: Suggests a repeated real solution, or one double root.
• Negative Discriminant: Means the solutions are complex or imaginary, thus not real in standard settings.

In the original exercise, we calculated the discriminant as 9:\[b^2 - 4ac = 1^2 - 4 \cdot 2 \cdot (-1) = 9\]A positive result confirmed that the equation \( 2u^2 + u - 1 = 0 \) had two distinct, real solutions, leading us to the corresponding \(\sin x\) values.