Problem 20 - Algebra and Trigonometry

Question

Find all solutions of the equation. $$2 \sin ^{2} x-\sin x-1=0$$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{\pi}{2} + 2k\pi\), \(x = \frac{7\pi}{6} + 2k\pi\), \(x = \frac{11\pi}{6} + 2k\pi\) for integer \(k\).

Step 1: Recognize the Quadratic Form

Notice that the equation \(2 \sin^2 x - \sin x - 1 = 0\) is in the form of a quadratic equation, where \(\sin x\) is the variable.

Step 2: Substitute Variables

Let \(y = \sin x\). Substitute \(y\) into the equation to get \(2y^2 - y - 1 = 0\).

Step 3: Solve the Quadratic Equation

Use the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 2\), \(b = -1\), and \(c = -1\). Calculate the discriminant: \((-1)^2 - 4 \times 2 \times (-1) = 1 + 8 = 9\). Thus, \(y = \frac{1 \pm \sqrt{9}}{4}\), giving the solutions \(y = 1\) and \(y = -\frac{1}{2}\).

Step 4: Determine Possible Angles for \(y = \sin x\)

For \(y = 1\), we know that \(\sin x = 1\) occurs at \(x = \frac{\pi}{2} + 2k\pi\), where \(k\) is any integer.

Step 5: Solve for Angles when \(y = -\frac{1}{2}\)

For \(y = -\frac{1}{2}\), \(\sin x = -\frac{1}{2}\) occurs at \(x = \frac{7\pi}{6} + 2k\pi\) and \(x = \frac{11\pi}{6} + 2k\pi\), where \(k\) is an integer.

Step 6: Write the General Solution

Combining results from Steps 4 and 5, the full solution set is: \(x = \frac{\pi}{2} + 2k\pi\), \(x = \frac{7\pi}{6} + 2k\pi\), \(x = \frac{11\pi}{6} + 2k\pi\) for any integer \(k\).

Key Concepts

Understanding the Quadratic Form

The equation \(2 \sin^2 x - \sin x - 1 = 0\) may look daunting at first, but if you observe closely, you'll recognize its quadratic nature. A quadratic equation is generally represented as \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) represents the variable. Here, our variable is \(\sin x\), which acts just like \(x\) in regular quadratic equations.

By substituting \(y = \sin x\), we transform the trigonometric equation into a purely algebraic one: \(2y^2 - y - 1 = 0\). This makes it straightforward to apply algebraic techniques, such as factoring or using the quadratic formula, to find the solutions.

Recognizing the quadratic form in trigonometric equations is a key skill in solving these types of equations. It allows you to utilize well-known algebraic methods to tackle problems involving the sine function.

The Versatile Sine Function

The sine function, denoted as \(\sin x\), is one of the most fundamental functions in trigonometry.

It measures the y-coordinate of a point on the unit circle that corresponds to a given angle \(x\).

• Values of sine range from -1 to 1, making it a periodic and bounded function.
• The sine function reaches its maximum value of 1 at \(x = \frac{\pi}{2} + 2k\pi\) and its minimum value of -1 at \(x = \frac{3\pi}{2} + 2k\pi\), where \(k\) is any integer capturing the periodicity.

In solving trigonometric equations like \(2 \sin^2 x - \sin x - 1 = 0\), understanding the periodic nature and range of the sine function is essential.

It helps in predicting possible angles or solutions and recognizing patterns or symmetries in the solutions.

Applying the Quadratic Formula

To solve quadratic equations such as \(2y^2 - y - 1 = 0\), the quadratic formula is extremely useful. This formula is given by
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a\), \(b\), and \(c\) are coefficients from the equation. It's a reliable method for finding the roots of any quadratic equation:

• First, calculate the discriminant \(b^2 - 4ac\), which determines the nature of the roots (real or complex).
• In our case, the discriminant \((-1)^2 - 4 \times 2 \times (-1) = 9\) is positive, so we have two real and distinct solutions.

Substitute back the values, and the formula yields roots \(y = 1\) and \(y = -\frac{1}{2}\).

The quadratic formula is invaluable for swiftly solving polynomial equations, especially when factoring isn't apparent or possible.

The General Solution of Trigonometric Equations

Once you find specific solutions to equations involving trigonometric functions, it's crucial to express them in a general form. This includes all possible occurrences of solutions given the periodic nature of trigonometric functions.

• For \(\sin x = 1\), solutions occur at \(x = \frac{\pi}{2} + 2k\pi\).
• For \(\sin x = -\frac{1}{2}\), solutions occur at \(x = \frac{7\pi}{6} + 2k\pi\) and \(x = \frac{11\pi}{6} + 2k\pi\).

Here, \(k\) is an integer representing how many times you 'cycle' through a full period of the sine function (which is \(2\pi\)).

The general solution combines all specific solutions into one unified expression, detailing every angle where the original equation holds true. Understanding this concept links algebraic solutions with their trigonometric contexts, providing a complete view of the problem.