Problem 71 - Calculus for Biology and Medicine

Question

Solve \(2 \cos \theta \sin \theta=\sin \theta\) on \([0,2 \pi)\)

Step-by-Step Solution

Answer

\(\theta = 0, \pi, \frac{\pi}{3}, \frac{5\pi}{3}\).

Step 1: Identify Common Factor

First, observe the equation \(2\cos\theta\sin\theta = \sin\theta\). We notice that both terms on either side of the equation contain \(\sin \theta\) as a factor. To simplify the expression, factor out \(\sin \theta\) from the left-hand side.

Step 2: Rewrite with Factoring

Factoring \(\sin \theta\) from the left-hand side, the equation becomes \(\sin \theta (2 \cos \theta - 1) = 0\). This setup allows us to solve for each factor separately.

Step 3: Solve for Each Factor Equating to Zero

Set each factor to zero and solve. First, \(\sin \theta = 0\), which gives solutions \(\theta = 0, \pi,\) since sine is zero at these points within the interval \([0, 2\pi)\). Next, solve \(2 \cos \theta - 1 = 0\), which simplifies to \(\cos \theta = \frac{1}{2}\).

Step 4: Solve \(\cos \theta = \frac{1}{2}\)

Find \(\theta\) such that \(\cos\theta = \frac{1}{2}\). This occurs at \(\theta = \frac{\pi}{3}\) and \(\theta = \frac{5\pi}{3}\) within the interval \([0, 2\pi)\).

Step 5: Combine All Solutions

Combine all found solutions. From the two cases: \(\theta = 0, \pi, \frac{\pi}{3}, \frac{5\pi}{3}\). These values are all within the interval \([0, 2\pi)\).

Key Concepts

Factoring in Equations

Factoring is a powerful technique used to simplify and solve equations. In trigonometric equations, just like in algebraic ones, factoring can make complex expressions more manageable by revealing simpler components. Let's explore how we factor in this type of problem.

The equation given in our problem was \(2\cos\theta\sin\theta = \sin\theta\). To apply factoring, we first look for a common term on both sides of the equation. Here, we noticed that \(\sin\theta\) appears in each term (left-side and right-side). This is our signal to factor it out.

By factoring out \(\sin\theta\), the equation transforms into \(\sin\theta (2\cos\theta - 1) = 0\). Breaking down a trigonometric equation into two parts using a factor simplifies finding solutions. Once you factor, you set each factor equal to zero resulting in simpler equations to solve.

Solving Trigonometric Identities

Solving trigonometric identities can initially seem intimidating, but breaking them down into achievable parts makes them manageable. After factoring in trigonometric equations, each part represents a trigonometric identity which is solved separately.

In our problem, once we had factored the equation to \(\sin\theta (2\cos\theta - 1) = 0\), we split it into two separate equations: \(\sin\theta = 0\) and \(2\cos\theta - 1 = 0\). Each equation has its own way to find solutions.

For \(\sin\theta = 0\), recognize that sine equals zero at specific angles, such as \(\theta = 0\) and \(\theta = \pi\) within our interval \([0, 2\pi)\). For the equation \(2\cos \theta = 1\), solve using basic algebra to find \(\cos \theta = \frac{1}{2}\). Solving these identities often involves knowing key trigonometric values or using an inverse trig function.

Unit Circle Solutions

The unit circle is an essential tool for solving trigonometric equations. It provides geometric insight into where and when certain trigonometric values occur.

In solving for \(cos \theta = \frac{1}{2}\), the unit circle is invaluable. The values of cosine and sine are often memorized at crucial angles on the unit circle. Cosine equals \(\frac{1}{2}\) at angles \(\theta = \frac{\pi}{3}\) and \(\theta = \frac{5\pi}{3}\) in the interval \([0, 2\pi)\).

The unit circle helps because it provides a direct reference for these angles and verifies where sine, cosine, and tangent achieve particular values. By using it, students can confidently determine solutions to trigonometric equations without solely relying on memorization. This way, the unit circle acts as both a practical tool and a visual aid in finding and confirming solutions.