Problem 98 - Algebra and Trigonometry

Question

Use the most appropriate method to solve each equation on the interval \([0,2 \pi) .\) Use exact values where possible or give approximate solutions correct to four decimal places. $$ 2 \sin 3 x+\sqrt{3}=0 $$

Step-by-Step Solution

Answer

The solutions to the equation \(2\sin 3x + \sqrt{3} = 0\) in the interval \([0,2\pi)\) are \(x = 4\pi/9, 5\pi/9, 8\pi/9, \pi\).

Step 1: Isolate the Trigonometric Term

Rearrange the equation to isolate the sine term: \(2 \sin 3x = - \sqrt{3}\), then divide both sides by 2 to isolate the sine expression: \(\sin 3x = - \sqrt{3}/2\).

Step 2: Find the Angles

The sin function is -\(\sqrt{3}/2\) at \(\pi/3\) and \(\pi\) in the unit circle. Here the period of function is double due to the factor 3 in front of x. Therefore, the solution to \(\sin 3x = - \sqrt{3}/2\) is \(3x = 4\pi/3, 5\pi/3, 2\pi + 4\pi/3, 2\pi + 5\pi/3\) etc, where the additional solutions come from the periodicity of sine function.

Step 3: Calculate the Value of x

Now divide 3 from the solutions obtained from step 2 to get the value of x in the given interval. Therefore, \(x = 4\pi/9, 5\pi/9, 8\pi/9, \pi\).

Key Concepts

Solutions in Intervals

In trigonometry, solving equations within a specific interval is a crucial skill. The interval \([0, 2\pi)\) often appears, especially in problems involving the unit circle. This interval is chosen because it precisely represents one full rotation around the circle, starting and ending just before \(2\pi\) radians.
For the equation \(2 \sin 3x + \sqrt{3} = 0\), isolating the sine function gives us \(\sin 3x = -\sqrt{3}/2\). The challenge is to find solutions for \(3x\) such that they fit within the interval for which their calculated value of \(x\) remains within \([0, 2\pi)\).
Consider that the coefficient \(3\) on \(x\) affects how we look at the interval: you need to solve it for \(3x\) first and then adjust to the allowable range for \(x\). This leads us to find solutions for \(x\) such as \((4\pi/9)\), \((5\pi/9)\), \((8\pi/9)\), and \(\pi\). Always ensure that those solutions lie in the specified interval. Continually checking into the interval ensures that all possible solutions are considered and captured.

Exact Values

When solving trigonometric equations, exact values are preferred for precision. In the given problem, the equation is simplified to \(\sin 3x = -\sqrt{3}/2\). These values are known as special angles, which can be found on the unit circle.
For the sine function, \(-\sqrt{3}/2\) corresponds to angles such as \(4\pi/3\) and \(5\pi/3\), angles which are derived from recognizing symmetry properties of the sine wave and the unit circle.
To derive the exact values efficiently, you need to know fundamental special angles and their trigonometric ratios. In this scenario, those angles are \(4\pi/3\) and \(5\pi/3\), which live in the third and fourth quadrants of a unit circle, providing solutions for \(3x\). These angles, divided by 3, translate back into solutions for \(x\) without needing decimal approximations.

Periodicity of Trigonometric Functions

The periodic nature of trigonometric functions like sine and cosine is essential in solving equations. They repeat their values at regular intervals, known as periods.
For sine and cosine, this period is typically \(2\pi\). However, multiplying \(x\) by a coefficient, such as the 3 in the problem’s \(3x\), alters this period. Consequently, the function's effective periods change to \(2\pi/3\), making the pattern of the function repeat every \(2\pi/3\) radians.
Understanding this concept is critical for identifying all possible solutions within a given interval. Since \(3x\) repeats its cycle every \(2\pi/3\) radians, accounting for this periodicity lets you uncover all angles meeting the equation's criteria. The periodicity lets us expand the set of solutions to be divided later by 3, simplifying into individual values for \(x\). Knowing how to handle these repetitive cycles ensures full coverage of all potential solutions in any specified range.