Problem 17 - A Graphical Approach to Precalculus with Limits

Question

Solve each equation for solutions over the interval \([0,2 \pi)\) by first solving for the trigonometric finction. Do not use a calculator. $$4(1+\sin x)(1-\sin x)=3$$

Step-by-Step Solution

Answer

The solutions are \( x = \frac{\pi}{6}, \frac{11\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6} \) within \([0, 2\pi)\).

Step 1: Expand the Trigonometric Equation

Start by expanding the left-hand expression using the difference of squares identity, \[(a-b)(a+b) = a^2 - b^2.\]Here, identify \( a = 1 \) and \( b = \sin x \), so \[ (1+ ext{sin} x)(1- ext{sin} x) = 1^2 - ( ext{sin} x)^2 = 1 - ext{sin}^2 x. \] Therefore, substitute into the equation:\[ 4(1- ext{sin}^2 x) = 3. \]

Step 2: Apply the Pythagorean Identity

Recognize that the Pythagorean identity states \[ ext{sin}^2 x + ext{cos}^2 x = 1. \] Substitute \( 1 - ext{sin}^2 x \) as \( ext{cos}^2 x \):\[ 4 ext{cos}^2 x = 3. \]

Step 3: Solve for \( \text{cos}^2 x \)

Divide both sides of the equation by 4:\[ \text{cos}^2 x = \frac{3}{4}. \]

Step 4: Solve for \( \text{cos} x \)

Take the square root of both sides:\[ \text{cos} x = \pm \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2}. \] This gives potential solutions for \( \text{cos} x \) as \( \text{cos} x = \frac{\sqrt{3}}{2} \) and \( \text{cos} x = -\frac{\sqrt{3}}{2}. \)

Step 5: Determine All Solutions in the Interval \([0, 2\pi)\)

For \( \text{cos} x = \frac{\sqrt{3}}{2} \), the angles \( x \) in the interval \([0, 2\pi)\) are \( x = \frac{\pi}{6} \) and \( x = \frac{11\pi}{6}. \)For \( \text{cos} x = -\frac{\sqrt{3}}{2} \), the angles \( x \) are \( x = \frac{5\pi}{6} \) and \( x = \frac{7\pi}{6}. \)

Key Concepts

Difference of Squares

The difference of squares is a fundamental algebraic concept where an expression is written as \((a-b)(a+b) = a^2 - b^2\). In this trigonometric equation, the expression \((1+\sin x)(1-\sin x)\) uses this concept. Here, we identify \(a = 1\) and \(b = \sin x\). By applying the difference of squares, the expression simplifies to \(1 - \sin^2 x\). This simplification is crucial as it transforms a complicated multiplication into a simple subtraction problem.
Such transformations are common in algebra and are essential tools to simplify and solve equations efficiently. Understanding this concept allows us to see how two potential multiplication components can be broken down into a subtraction of squares, easing the way forward in solving the equation.

Pythagorean Identity

The Pythagorean Identity is one of the cornerstone identities in trigonometry: \[\sin^2 x + \cos^2 x = 1.\] It speaks to the intrinsic relationship between sine and cosine functions. In this application, by substituting \(1 - \sin^2 x\) with \(\cos^2 x\) derived from the Pythagorean identity, we rebalance the equation allowing us to solve for \(\cos x\).

• It is essential to recognize this identity promptly as it aids massively in simplifying trigonometric equations.
• Its application reveals solutions to problems which might initially appear complicated.

Mastering the Pythagorean identity empowers students to maneuver through trigonometric expressions with ease, revealing elegant solutions.

Unit Circle

The unit circle is a vital tool in trigonometry that connects angles and lengths back to a circle with radius one centered at the origin of a coordinate plane. On this circle:

• The \(x\)-coordinate of a point is the cosine of the angle.
• The \(y\)-coordinate is the sine of the angle.

In solving for \(\cos x = \pm \frac{\sqrt{3}}{2}\), we draw upon the unit circle. We identify specific angles where cosine takes these values by remembering standard angles measured in radians.
The unit circle ratifies exact values at these critical angles which help trace our calculated cosine solutions back to tangible angle measures on the circle. This connection among coordinate length, angle, and calculations is crucial for exact solution determination.

Solution Interval

In trigonometry, solving an equation often involves finding solutions over specific intervals, commonly \([0, 2\pi)\) for one complete cycle of a periodic function. This range represents all possible angle values in radians for one full circle.

• As trigonometric functions are periodic, multiple solutions can occur inside this interval.
• Recognizing solutions in given intervals demands clear understanding of when and how trigonometric values repeat.

In this problem, for the values \(\cos x = \frac{\sqrt{3}}{2}\) and \(\cos x = -\frac{\sqrt{3}}{2}\), we identified the corresponding angles: \(x = \frac{\pi}{6}, \frac{11\pi}{6}, \frac{5\pi}{6},\) and \(\frac{7\pi}{6}\). Each pinpointed angle falls within the specified interval, offering a comprehensive list of solutions. That's why knowing the interval helps establish a clear boundary for solving trigonometric equations.