Problem 26 - Algebra and Trigonometry

Question

Find all solutions of the equation. $$2 \cos 2 x+1=0$$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{\pi}{3} + k\pi\) and \(x = \frac{2\pi}{3} + k\pi\) for integer \(k\).

Step 1: Isolate Cosine Expression

The given equation is \( 2 \cos 2x + 1 = 0 \). First, we need to isolate the cosine expression. Subtract 1 from both sides to get: \[ 2 \cos 2x = -1 \]. Next, divide both sides by 2: \[ \cos 2x = -\frac{1}{2} \].

Step 2: Solve for the Angle

We have \( \cos 2x = -\frac{1}{2} \). The cosine of an angle is \(-\frac{1}{2}\) at the angles \(2x = \frac{2\pi}{3} + 2k\pi\) and \(2x = \frac{4\pi}{3} + 2k\pi\), where \(k\) is an integer, representing the general solution for cosine.

Step 3: Solve for x

Now, solve for \(x\) from the equations we found: \(2x = \frac{2\pi}{3} + 2k\pi\) leads to \(x = \frac{\pi}{3} + k\pi\), and \(2x = \frac{4\pi}{3} + 2k\pi\) leads to \(x = \frac{2\pi}{3} + k\pi\). These are the general solutions for \(x\).

Key Concepts

Cosine Function

The cosine function, often represented as \( \cos \theta \), is a fundamental trigonometric function that relates the angle \( \theta \) to the ratio of the adjacent side of a right triangle to its hypotenuse. Within the unit circle framework, cosine corresponds to the x-coordinate of a point located on the circle's circumference.
When analyzing trigonometric equations like \( 2 \cos 2x + 1 = 0 \), understanding the behavior of the cosine function is crucial. The cosine function is periodic, with a standard period of \(2\pi\), meaning it repeats its values every \(2\pi\) radians.

• This periodicity allows the cosine function to have the same value at multiple points over its domain.
• This is why multiple angles can satisfy the same cosine value within one cycle.

In our exercise, recognizing that \( \cos 2x = -\frac{1}{2} \) implies that we are working with specific angles where the cosine function takes on that value. This understanding aids in finding particular solutions within the typical range \([-\pi, \pi]\), before considering the general solutions.

Angle Solving

Solving for angles in trigonometric equations, such as \( \cos 2x = -\frac{1}{2} \), requires knowledge of the unit circle and several key angles. The equation asks us to determine which angles 2x equates to, where the cosine value is \(-\frac{1}{2}\).
For the cosine function, this specifically occurs at \(\frac{2\pi}{3}\) and \(\frac{4\pi}{3}\) radians within one cycle of the circle.

• These angles represent the standard solutions when \(x\) is limited to \([-\pi, \pi]\).
• Each corresponding angle marks positions where the horizontal coordinate is negative, matching \(-\frac{1}{2}\).

To solve \(2x = \frac{2\pi}{3} + 2k\pi\) and \(2x = \frac{4\pi}{3} + 2k\pi\), we divide by 2, deriving expressions for \(x\) as \(x = \frac{\pi}{3} + k\pi\) and \(x = \frac{2\pi}{3} + k\pi\). This approach systematically considers all possible angle solutions through multiples of \(\pi\).

General Solution

The general solution of a trigonometric equation captures all possible solutions across its infinite, periodic nature. For an equation like \( \cos 2x = -\frac{1}{2} \), it’s essential to provide a solution format that encompasses every instance where the function returns this value.
Cosine's periodicity demands that solutions not only capture angles in a single traversal of the unit circle but also loop infinitely along its entire domain.

• This is accomplished using integers \(k\), indicating multiple cycles in the solution.
• For our specific example, the general solutions are \(x = \frac{\pi}{3} + k\pi\) and \(x = \frac{2\pi}{3} + k\pi\).
• The parameter \(k\) is an integer that accounts for these infinite cycles.

Recognizing the generality of solution sets is crucial to mastering trigonometric equations, ensuring that all potential angle measures are included, no matter how many times the system repeats.