Problem 11 - A Graphical Approach to Precalculus with Limits

Question

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

Step-by-Step Solution

Answer

Solutions: \(x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{\pi}{6}, \frac{5\pi}{6}\).

Step 1: Use a Double Angle Identity

First, recognize the double angle identity: \( \sin 2x = 2 \sin x \cos x \). Substituting this into the equation, we have: \[ 2 \sin x \cos x - \cos x = 0 \].

Step 2: Factor Out Common Terms

Now, factor out \( \cos x \) from the equation: \[ \cos x (2 \sin x - 1) = 0 \]. This gives two equations to solve: \[ \cos x = 0 \] and \[ 2 \sin x - 1 = 0 \].

Step 3: Solve the First Equation

Solve \( \cos x = 0 \). In the interval \([0, 2\pi)\), \( \cos x = 0 \) at:\[ x = \frac{\pi}{2}, \frac{3\pi}{2} \].

Step 4: Solve the Second Equation

Solve \( 2 \sin x = 1 \) by dividing both sides by 2: \[ \sin x = \frac{1}{2} \].In the interval \([0, 2\pi)\), \( \sin x = \frac{1}{2} \) at:\[ x = \frac{\pi}{6}, \frac{5\pi}{6} \].

Step 5: List All Solutions

Combine all the solutions found from the previous steps, which are:\[ x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{\pi}{6}, \frac{5\pi}{6} \].These are all the values of \( x \) in the interval \([0, 2\pi)\) that satisfy the original equation.

Key Concepts

Double Angle Identities

In trigonometry, double angle identities are powerful tools used to simplify expressions involving trigonometric functions. Understanding these identities can hugely benefit in solving complex trigonometric equations.
💡 To recall, a double angle identity for sine is expressed as:

• \( \sin 2x = 2 \sin x \cos x \)

This identity allows us to turn the double angle \( 2x \) into a single angle expression, making it more manageable. In our problem, replacing \( \sin 2x \) with \( 2 \sin x \cos x \) transformed the equation from \( \sin 2x - \cos x = 0 \) to \( 2 \sin x \cos x - \cos x = 0 \).
As seen, utilizing this double angle formula simplifies the equation, leading to easier factorization. Double angle identities are especially useful in turning products and sums of trigonometric functions into forms that are more straightforward to solve.

Factorization

Factorization, in the context of solving equations, is a method used to simplify expressions by extracting common terms. This process allows for breaking down an equation into simpler factors which can be solved individually.
In the given trigonometric equation, \( 2 \sin x \cos x - \cos x = 0 \), we can see that \( \cos x \) is a common factor in both terms. Factoring \( \cos x \) out, the equation becomes:

• \( \cos x (2 \sin x - 1) = 0 \)

Through factorization, we split the equation into two separate equations: \( \cos x = 0 \) and \( 2 \sin x - 1 = 0 \).
💡 Each of these smaller equations can be solved independently. Factorization reveals solutions that might not be obvious with the original equation. Understanding this process helps in identifying multiple solutions effectively.

Interval Notation

Interval notation is a way of representing the set of solutions for equations or inequalities. It's particularly useful in precise communication of which parts of the number line are included in a solution.
The interval \([0, 2\pi)\) specifies all possible values for \( x \) in this problem, indicating solutions must be greater than or equal to \(0\) and less than \(2\pi\).
After solving the individual trigonometric equations, we find solutions:

• For \( \cos x = 0 \): \( x = \frac{\pi}{2}, \frac{3\pi}{2} \).
• For \( \sin x = \frac{1}{2} \): \( x = \frac{\pi}{6}, \frac{5\pi}{6} \).

These results are compiled to list all valid solutions within the specified interval: \( x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{\pi}{6}, \frac{5\pi}{6} \). Interval notation ensures clarity in denoting range and inclusion, preventing ambiguity.