Problem 3 - 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. $$5 \sin x-6=0$$

Step-by-Step Solution

Answer

There are no solutions in the interval \([0, 2\pi)\) as \(\sin x = \frac{6}{5}\) is not possible.

Step 1: Isolate the Trigonometric Function

First, we need to solve the given equation for the trigonometric function. The equation is \( 5 \sin x - 6 = 0 \). To isolate \( \sin x \), add 6 to both sides:\[ 5 \sin x = 6 \]

Step 2: Solve for Sine

Now, divide both sides by 5 to solve for \( \sin x \):\[ \sin x = \frac{6}{5} \] Notice that \( \frac{6}{5} \) is greater than 1. The sine of an angle in a real number interval \( [0, 2\pi) \) can't exceed 1, as the sine function's range is from -1 to 1. Therefore, there is no solution for \( \sin x = \frac{6}{5} \) in this interval.

Step 3: Conclude with No Solution

Since the value \( \sin x = \frac{6}{5} \) falls outside the valid range for the sine function (i.e., greater than 1), there are no solutions to the equation \( 5 \sin x - 6 = 0 \) in the interval \( [0, 2\pi) \).

Key Concepts

Understanding the Sine Function

The sine function is a fundamental part of trigonometry, representing one of the basic trigonometric ratios derived from a right triangle. It relates the ratio of the length of the side opposite an angle to the length of the hypotenuse. In a unit circle, which is a circle with a radius of 1, the sine of an angle \(x\) corresponds to the y-coordinate of the point on the circle that intersects with the given angle starting from the positive x-axis.
The sine function has a range from -1 to 1. This means that any sine value must fall within this interval. If we encounter a situation where the sine value exceeds this range, as seen in the given exercise where \(\sin x = \frac{6}{5}\) (which is greater than 1), it indicates that there's a misunderstanding or miscalculation, as such values are not possible for real angles. Understanding this limitation is key to solving trigonometric equations effectively.
Furthermore, the sine function is periodic with a period of \(2\pi\), meaning it repeats its values every \(2\pi\) radians. Thus, while solving equations, this periodic nature allows us to consider only one cycle, typically taken as the interval \( [0, 2\pi) \).

Intervals in Trigonometry

Intervals in trigonometry are essential for specifying the domain within which we're looking to solve equations. For example, in the problem, solutions are sought over the interval \( [0, 2\pi) \). This interval represents one full rotation around a circle, relevant in explaining periodic functions like sine and cosine.
It's important to clearly understand the notation and meaning of intervals. The bracket \( [ \) indicates inclusion of the boundary, so 0 is part of the interval, whereas the parenthesis \( ) \) signifies exclusion, meaning \(...2\pi\) isn't included. This distinction ensures we consider only angles within this specified range, avoiding ambiguities like identifying angles greater than \(2\pi\) or less than 0.

• When solving trigonometric equations, ensure the angle solutions lie within the given interval.
• Recognize that intervals help capture the cyclical nature of trig functions.

Knowing how intervals work aids in proper application of trigonometric functions to real-world problems, providing accurate solutions.

Solving Equations in Trigonometry

Solving trigonometric equations involves finding angles that satisfy given conditions using trigonometric functions. In our initial equation \(5 \sin x - 6 = 0\), the procedure involved isolating the sine function first. Steps often involve:

• Rearranging the equation to find the trigonometric function, here \(\sin x\).
• Verifying if the trigonometric value is within its possible range.

If the sine value doesn't fall between -1 and 1, as in \(\sin x = \frac{6}{5}\), no real angles exist for the equation within typical real number intervals. This lack of solution emphasizes the sine function's range limit.Solving trigonometric equations often requires understanding not only the immediate mathematical manipulation but also concept enforcement like:

• Knowing function ranges to identify viable solutions.
• Understanding periodic nature of functions to utilize appropriate intervals effectively.
• Checking angles within specified domains to meet all given conditions accurately.

Mastery of these steps ensures strong problem-solving skills in trigonometry, equipping students to tackle complex real-world problems efficiently.