Problem 2 - A Graphical Approach to Precalculus with Limits

Question

Solve each equation for solutions over the interval \([0,2 \pi)\) by first solving for the trigonometric finction. Do not use a calculator. $$2 \sin x+1=0$$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{7\pi}{6} \) and \(x = \frac{11\pi}{6} \).

Step 1: Isolate the Trigonometric Term

Start with the given equation: \( 2\sin x + 1 = 0 \). Subtract \(1\) from both sides to isolate the \(2\sin x\) term:\[ 2\sin x = -1 \]

Step 2: Solve for Sine Function

Divide each side of the equation by 2 to solve for \(\sin x\):\[ \sin x = -\frac{1}{2} \]

Step 3: Determine the Relevant Angles

Identify where \( \sin x = -\frac{1}{2} \) on the unit circle. Since sine is negative, we check the third and fourth quadrants where sine is negative. The reference angle for \( \sin x = \frac{1}{2} \) is \( \frac{\pi}{6} \). Thus, in the third quadrant, the angle will be \( \pi + \frac{\pi}{6} = \frac{7\pi}{6} \) and in the fourth quadrant, it will be \( 2\pi - \frac{\pi}{6} = \frac{11\pi}{6} \).

Step 4: Verify Solutions within Interval

Verify that the solutions \(x = \frac{7\pi}{6} \) and \(x = \frac{11\pi}{6} \) both lie within the interval \([0, 2\pi)\). Both values are within this interval, confirming they are valid solutions.

Key Concepts

Unit Circle

The unit circle is an essential concept in trigonometry, helping us solve trigonometric equations like the one given. The circle has a radius of 1 and is centered at the origin of a coordinate plane (0,0). The unit circle allows us to find the values of trigonometric functions for various angles.

In the context of our exercise, the unit circle helps determine where the sine function reaches a value of \(-\frac{1}{2}\).

• Angles on the unit circle are measured in radians, which is often the preferred unit over degrees in mathematics.
• The sine of an angle corresponds to the y-coordinate of its endpoint when the angle is placed in standard position.
• In the unit circle, positive sine values are found in the first and second quadrants, while negative sine values are in the third and fourth quadrants.

Using the circle, we can see that our angle solutions fall in the third and fourth quadrants, where sine takes negative values.

Sine Function

The sine function is a fundamental component of trigonometry. It represents the ratio of the opposite side to the hypotenuse in a right triangle. In terms of the unit circle, it gives the y-coordinate of a point on the circle.

To solve the equation \( \sin x = -\frac{1}{2} \), we look for angles on the unit circle that correspond to this sine value.

• The sine function is periodic with a period of \(2\pi\), meaning that it repeats every \(2\pi\) radians.
• Sine values range from -1 to 1.
• In one full period, the sine of an angle is zero at \(0, \pi\) and \(2\pi\), peaks at 1 at \(\frac{\pi}{2}\), and reaches -1 at \(\frac{3\pi}{2}\)

For the given exercise, using the sine function's property that it is negative in the third and fourth quadrants helps us pinpoint where \( \sin x = -\frac{1}{2} \), simplifying the process of finding corresponding angles.

Reference Angle

The concept of a reference angle is crucial when solving trigonometric equations without a calculator. A reference angle is the smallest angle that a given angle makes with the x-axis. It is always positive and typically found by working within the first quadrant.

For the equation \( \sin x = -\frac{1}{2} \), the corresponding positive reference angle is \( \frac{\pi}{6} \).

• The reference angle for sine is linked to the absolute value of the sine function.
• By understanding the reference angle, we can find where the function's value is the same in other quadrants.
• In this exercise, \( \sin x = \frac{1}{2} \) corresponds to \(x = \frac{\pi}{6}\) in the first quadrant, hence the third quadrant angle is \( \pi + \frac{\pi}{6} \) and the fourth quadrant one is \( 2\pi - \frac{\pi}{6} \).

Grasping the concept of reference angles helps find precise solutions easily by applying them in the appropriate quadrants.