Problem 64 - Precalculus Essentials

Question

Use an identity to solve each equation on the interval \([0,2 \pi)\) $$2 \cos ^{2} x-\sin x-1=0$$

Step-by-Step Solution

Answer

The solutions for the equation on the interval [0,2π) are \(x = \pi, \pi/6, 5\pi/6\).

Step 1: Substitute for \(\cos^{2} x\)

Substitute \(2\cos^{2} x\) with \(2 - 2\sin^{2} x\) using the identity \(\cos^{2} x = 1 - \sin^{2} x\). The equation transforms into: \(2 - 2\sin^{2} x - \sin x - 1 = 0\) which simplifies to: \(2\sin^{2} x + \sin x - 1 = 0\).

Step 2: Solve the Quadratic

The equation \(( 2\sin^{2} x + \sin x - 1 = 0)\) is now a quadratic equation in \(\sin x\). Using the quadratic formula \(\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}\), where \(a = 2\), \(b = 1\), and \(c = -1\), we find two solutions for \(\sin x\): (-1) and (0.5).

Step 3: Calculate \(x\)

Finding the angles \(x\) that correspond to \(\sin x = -1\) and \(\sin x = 0.5\), we have \(x = \pi + 2\pi m\) for \(m \in Z\) or \(x = \pi/6 + 2\pi n\) and \(x = 5\pi/6 + 2\pi n\) for \(n \in Z\). Since the interval is \([0, 2\pi]\), we discard any results outside this range. Therefore, the solutions are \(x = \pi, \pi/6, 5\pi/6\).

Key Concepts

Quadratic Equations

Quadratic equations play a vital role in solving trigonometric identities when they appear in trigonometric forms. In this context, a quadratic equation indicates an equation of the form \( ax^2 + bx + c = 0 \). A common tool for solving such equations is the quadratic formula:

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

This formula allows us to find the solutions for \( x \) by substituting the coefficients \( a \), \( b \), and \( c \) appropriately. In the exercise example, by identifying \( a = 2 \), \( b = 1 \), and \( c = -1 \), we plug these values into the quadratic formula to find the potential solutions for \( \sin x \). The results we get after calculation can help unveil the solutions required for trigonometric functions because they often translate into specific values for sine or cosine functions, leading us directly to potential angle solutions.
This conversion from trigonometric terms into a known quadratic equation form is key to applying more familiar algebraic methods. It bridges the often complex-seeming gap between trigonometry and algebra.

Unit Circle

The unit circle is an essential tool in trigonometry, especially when it comes to linking angles with their sine and cosine values. The circle has a radius of 1, centered at the origin of a coordinate system. By moving around this circle, different angles can be represented as points along its circumference. These points allow us to determine sine and cosine values easily.

• On the unit circle, the x-coordinate corresponds to \( \cos \theta \), while the y-coordinate corresponds to \( \sin \theta \).
• A full rotation around the circle covers an interval \([0, 2\pi)\).

In our exercise, determining when \( \sin x = -1 \) or \( \sin x = 0.5 \) means finding angles, or positions, on the unit circle with these y-coordinate values. For instance, \( \sin x = -1 \) leads us to the angle \( \pi \), which is at the bottom of the circle, while \( \sin x = 0.5 \) corresponds to the angles \( \pi/6 \) and \( 5\pi/6 \). These positions represent the solutions given by the unit circle, calculated over the specified interval.

Sin and Cos Functions

Sine and cosine functions are fundamental trigonometric functions that describe the relationship between an angle of a right triangle and the lengths of its sides. They also relate to the unit circle and are periodic with a cycle of \(2\pi \). Important properties of the sine and cosine functions are:

• \(\sin(0) = 0\), \(\cos(0) = 1\)
• Sine function is odd: \(\sin(-\theta) = -\sin(\theta)\)
• Cosine function is even: \(\cos(-\theta) = \cos(\theta)\)

In solving trigonometric equations, we frequently use identities like \(\cos^2 x = 1 - \sin^2 x\) to transform and solve the equations. These transformations help in expressing the equation in terms of a single function (like sine in this case) so that algebraic methods can be applied.
In the given exercise, we started with a mix of \( \cos^2 x \) and \( \sin x \). By using the identity mentioned, we rewrote \( \cos^2 x \) to convert everything into sine terms. This allows simplifying the equation into a quadratic form that was easier to solve. Recognizing these transformations and understanding where they come from is vital for decomposing and solving trigonometric problems.