91Ó°ÊÓ

Quadratic equations are polynomial equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(a eq 0\). In trigonometry, quadratic-like equations can involve trigonometric functions such as cosine instead of a simple variable like \(x\).

In the given problem, the expression \(4 \cos^2 x - 4 \cos x + 1 = 0\) appears similar to the standard quadratic form. By substituting \(u = \cos x\), the equation transforms into \(4u^2 - 4u + 1 = 0\), which is now a quadratic equation in the variable \(u\).

To solve for \(u\), you can use the quadratic formula:

• The formula is \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
• Plug in \(a = 4\), \(b = -4\), and \(c = 1\) to find the value of \(u\).

This technique is helpful in solving trigonometric equations by reducing them to algebraic ones, making it much easier to manage and solve.

The cosine function, written as \(\cos x\), is a fundamental trigonometric function that reflects the ratio of the adjacent side to the hypotenuse in a right-angled triangle. It is also crucial in analyzing periodic phenomena and forms a wave-like pattern when graphed.

In this specific exercise, identifying \(\cos x = \frac{1}{2}\) as a solution was key in resolving the trigonometric portion of the problem. The cosine function is periodic with a period of \(2\pi\), meaning it repeats its values every \(2\pi\) radians.

When solving \(\cos x = \frac{1}{2}\), it's essential to consider the standard angles which yield this result:

• First quadrant: \(x = \frac{\pi}{3}\)
• Fourth quadrant: \(x = \frac{5\pi}{3}\)

Each of these solutions can be extended with the full period \(2k\pi\) to cover all possible solutions.

The unit circle is a crucial concept in trigonometry, providing a geometrical way to visualize trigonometric functions. It is a circle with a radius of exactly one, centered at the origin of a coordinate plane. The angle \(x\) is measured from the positive x-axis, and for any angle \(x\), the point on the unit circle is \((\cos x, \sin x)\).

In this problem, finding where \(\cos x = \frac{1}{2}\) means identifying points on the unit circle where the x-coordinate (cosine value) is \(\frac{1}{2}\).

• This occurs at \(\frac{\pi}{3}\) in the first quadrant.
• It also occurs at \(\frac{5\pi}{3}\) in the fourth quadrant.

These results stem from understanding the symmetrical properties of the unit circle, where the cosine is positive both in the first and the fourth quadrants.