Problem 60 - Algebra and Trigonometry

Question

Solve each equation on the interval \([0,2 \pi)\) $$ \cos x-2 \sin x \cos x=0 $$

Step-by-Step Solution

Answer

The solutions to the equation are \( x = 0, \pi/3, 2\pi/3, \pi, 4\pi/3 \), and \( 5\pi/3 \) on the interval \([0,2 \pi)\)

Step 1: Identify and Obtain the Equations

The given equation is \( \cos x - 2 \sin x \cos x = 0 \). The task is to solve for \( x \). This equation needs to be simplified so that we can solve it.

Step 2: Apply Double Angle Identity

Notice that \( -2 \sin x \cos x \) is part of a double angle identity for sine: \( \sin 2x = 2 \sin x \cos x \). Using this, the equation can be rewritten as \( \cos x - \sin 2x = 0 \)

Step 3: Isolate the Value of \( x \)

Now, rearrange the equation to get \( \cos x = \sin 2x \). From the definitions of sine and cosine, we know that \( \sin( \pi/2 - x ) = \cos x \). So we can replace \( \cos x \) with \( \sin( \pi/2 - x ) \), and our equation becomes \( \sin( \pi/2 - x ) = \sin 2x \)

Step 4: Solve for \( x \)

We now have \( \sin( \pi/2 - x ) = \sin 2x \), and according to the equality properties of trigonometric functions, if \( \sin a = \sin b \), then \( a = n \pi + (-1)^n b \), where \( n \) is any integer. Applying this here gives us two sequences of possible solutions: \( \pi/2 - x = 2\pi n - 2x \) and \( \pi/2 - x = 2\pi n + 2x \). Solving these gives us \( x = \pi n \) and \( x = \pi n / 3 \) respectively, for \( n \) an integer.

Step 5: Find the Solutions in the Required Interval

We're only interested in solutions \( x \) where \( 0 \leq x < 2\pi \). For \( x = \pi n \), in this range \( n \) can be 0, 1, or 2 yielding \( x \) as \( 0, \pi, 2\pi \). For \( x = \pi n / 3 \), \( n = 1, 2, \ldots, 5 \) and our solutions are \( x = \pi/3, 2\pi/3, \pi, 4\pi/3, 5\pi/3 \). The solutions to the equation in the given range are therefore \( x = 0, \pi/3, 2\pi/3, \pi, 4\pi/3 \), and \( 5\pi/3 \).

Key Concepts

Double Angle Identities

Double angle identities are essential tools in trigonometry that help simplify and solve trigonometric equations. These identities express trigonometric functions of double angles (such as \(2x\)) in terms of single angles (like \(x\)). For instance, the double angle identity for sine is given by:

• \(\sin 2x = 2 \sin x \cos x\)

This means that instead of working with \(\sin 2x\), it can be replaced with \(2 \sin x \cos x\) in equations. This identity is useful when you see terms like \(\sin x \cos x\) in your equation. It helps in reducing complex expressions and makes solving the equations more straightforward.
In solving our original equation \(\cos x - 2 \sin x \cos x = 0\), the term \(-2 \sin x \cos x\) was recognized as part of the \(\sin 2x\) identity. Hence, the equation was rewritten as \(\cos x - \sin 2x = 0\). This simplification paved the way for finding the solutions more efficiently.

Sine and Cosine Equations

Sine and cosine equations are equations that feature the sine or cosine of an angle, or sometimes both. When dealing with these equations, it is often useful to employ identities and properties that relate sine and cosine to one another.
In the specific case of our equation, after applying the double angle identity, we derived that:

• \(\cos x = \sin 2x\)

To solve this, we utilize another relationship: \(\cos x = \sin(\pi/2 - x)\). This recognition allows rewriting the equation further to:

• \(\sin(\pi/2 - x) = \sin 2x\)

Given this, we use the property that if \(\sin a = \sin b\), then it leads to two primary scenarios:

• \(a = b + 2\pi n\)
• \(a = \pi - b + 2\pi n\)

These properties originate from the periodic nature of the sine function, which repeats every \(2\pi\). Solving such equations involves considering these general forms and finding values of \(x\) within specific intervals.

Solutions in Specific Intervals

When solving trigonometric equations, it's crucial to find solutions within a specified interval. In our problem, the interval was \([0, 2\pi)\). This means that any solution must fall between \(0\) and \(2\pi\), including \(0\) but not \(2\pi\).
During the solution process, you find lists of potential solutions, such as \(x = \pi n\) and \(x = \pi n / 3\) for integer values of \(n\). However, not all values fit the interval requirements. Thus, you must evaluate which specific \(n\) values yield solutions within the interval by:

• Setting \(n = 0, 1, 2, \ldots\)
• Calculating the corresponding \(x\) for each \(n\)
• Ensuring calculated \(x\) values lie in \(0 \leq x < 2\pi\)

By doing this careful checking, you ensure that all retained solutions are valid for the problem's constraints. In our example, the accepted solutions were \(x = 0, \pi/3, 2\pi/3, \pi, 4\pi/3,\) and \(5\pi/3\). Ensuring solutions are within the interval is a vital step to accurately solving trigonometric equations.