91Ó°ÊÓ

The cosine function, denoted as \( \cos(x) \), is a fundamental concept in trigonometry. It is one of the primary functions used to relate the angles and sides of a right triangle. The cosine function maps an angle \( x \) to the x-coordinate of a point on the unit circle, making it periodic with a cycle of \( 2\pi \) radians (or 360 degrees). This periodic nature also means \( \cos(x) \) repeats its values in a regular fashion, ranging from -1 to 1.

• At \( x = 0 \), \( \cos(0) = 1 \).
• At \( x = \frac{\pi}{2} \), \( \cos(\frac{\pi}{2}) = 0 \).
• At \( x = \pi \), \( \cos(\pi) = -1 \).
• At \( x = \frac{3\pi}{2} \), \( \cos(\frac{3\pi}{2}) = 0 \).
• Completing the cycle at \( x = 2\pi \), \( \cos(2\pi) = 1 \) again.

Understanding these key points and the behavior of the cosine function helps in solving trigonometric equations like the one in the exercise by recognizing patterns and possible solutions.

In trigonometry, identities are equations that hold true for all values of the included variables. One essential identity is the double angle formula for cosine:

\[ \cos(2x) = 2\cos^2(x) - 1 \]
This identity allows us to express cosine of a double angle (\(2x\)) in terms of the square of the cosine of the original angle (\(x\)).

Our original problem utilizes this identity to transform \( \cos(2x) \) into a quadratic form. Using identities like these can often simplify equations and make solving them more manageable. By applying this identity, what initially appears to be a complex trigonometric equation can potentially be simplified to a format that is easier to solve, such as a quadratic equation.

A quadratic equation is an equation of the second degree, generally in the form \( ax^2 + bx + c = 0 \). In this exercise, our equation appeared as:

\[ 2\cos^2(x) + 5\cos(x) - 3 = 0 \]
Here, \( \cos(x) \) is treated like a variable, making it a quadratic in \( \cos(x) \). Quadratic equations can be solved using various methods such as factoring, completing the square, or most commonly the quadratic formula:

\[ \cos(x) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For our equation, substituting \( a = 2 \), \( b = 5 \), and \( c = -3 \) into the quadratic formula yields solutions for \( \cos(x) \). This step transforms a trigonometric problem into a familiar algebraic one, which can then be treated using algebraic techniques.

Exact solutions refer to precise values that satisfy a mathematical equation, devoid of approximation. In trigonometry, exact solutions often relate to angles with known trigonometric values, such as \( \frac{\pi}{3}, \frac{\pi}{2}, \pi, \), and others. For the cosine equation with the quadratic roots, we found:

• \( \cos(x) = \frac{1}{2} \), with exact solutions \( x = \frac{\pi}{3} \) and \( x = \frac{5\pi}{3} \).
• \( \cos(x) = -3 \) was discarded since \( \cos(x) \) must be in the range \([-1, 1]\).

Solving for \( x \) when \( \cos(x) = \frac{1}{2} \), we utilize knowledge of the unit circle, where these angles occur naturally. This is what ensures the answers are exact — they conform precisely to the constraints of trigonometric functions and their symmetries around the unit circle. This approach not only provides mathematic accuracy but also lays the groundwork for understanding intrinsic trigonometric relationships.