91Ó°ÊÓ

A quadratic equation is a second-order polynomial equation in a single variable. It generally takes the form:

• \(ax^2 + bx + c = 0\)

The coefficients are real numbers, where \(a \, eq 0\). The solution is found using the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

The term inside the square root, \(b^2 - 4ac\), is called the discriminant. It determines the nature of the roots:

• Positive discriminant: two real roots
• Zero discriminant: one real root (repeated)
• Negative discriminant: no real roots (complex roots)

In the original exercise, we identified \(\sin\theta\) as the variable and solved the quadratic equation \(2\sin^2\theta - \sin\theta - 1 = 0\). We used the quadratic formula to find the solutions \(\sin\theta = 1\) and \(\sin\theta = -\frac{1}{2}\).

Trigonometric functions relate the angles of a triangle to the lengths of its sides. The most commonly used trigonometric functions are:

• Sine (\(\sin\)) - gives the ratio of the length of the opposite side to the hypotenuse
• Cosine (\(\cos\)) - gives the ratio of the length of the adjacent side to the hypotenuse
• Tangent (\(\tan\)) - gives the ratio of the length of the opposite side to the adjacent side

These functions are vital in solving trigonometric equations. For the given problem, we needed to find values of \(\theta\) where \(\sin\theta = 1\) and \(\sin\theta = -\frac{1}{2}\) within the interval \(0 \leq \theta < 2\pi\).

• For \(\sin\theta = 1\), \(\theta = \frac{\pi}{2}\)
• For \(\sin\theta = -\frac{1}{2}\), \(\theta = \frac{7\pi}{6}\) and \(\theta = \frac{11\pi}{6}\)

The unit circle is a circle with a radius of 1, centered at the origin of the coordinate plane. It's a powerful tool to visualize and understand trigonometric functions. Points on the unit circle can be described using coordinates (\(\cos(\theta)\), \(\sin(\theta)\)), where \(\theta\) is the angle formed with the positive x-axis.
The unit circle is divided into four quadrants, with specific angles providing simple trigonometric values:

• Quadrant I: \(0 \leq \theta < \frac{\pi}{2}\)
• Quadrant II: \(\frac{\pi}{2} \leq \theta < \pi\)
• Quadrant III: \(\pi \leq \theta < \frac{3\pi}{2}\)
• Quadrant IV: \(\frac{3\pi}{2} \leq \theta < 2\pi\)

For the given equation, these quadrants help determine the angles where \(\sin\theta = 1\) and \(\sin\theta = -\frac{1}{2}\). The corresponding angles on the unit circle are:

• \(\sin\theta = 1\) at \(\theta = \frac{\pi}{2}\) in Quadrant I
• \(\sin\theta = -\frac{1}{2}\) at \(\theta = \frac{7\pi}{6}\) in Quadrant III and \(\theta = \frac{11\pi}{6}\) in Quadrant IV

Using the unit circle simplifies finding solutions to trigonometric equations.