Problem 59 - Precalculus Essentials

Question

Solve the equation on the interval \([0,2 \pi)\) $$\sin x+2 \sin x \cos x=0$$

Step-by-Step Solution

Answer

The solutions to the equation on the interval [0, \(2\pi)\) are \(x = \frac{2\pi}{3}\) and \(x = \frac{4\pi}{3}\).

Step 1: Apply Trigonometric Identity

Replace \(2\sin{x}\cos{x}\) with \(\sin{2x}\). The equation is now \(\sin{x} + \sin{2x} = 0\). That simplifies the equation by removing a cosine term.

Step 2: Isolate \(\sin{x}\)

Rearrange the equation to isolate \(\sin{x}\): \(\sin{x} = -\sin{2x}\). This equation implies that \(x\) and \(2x\) must be opposite in value (since \(\sin{x}\) is positive and \(\sin{-x}\) is negative).

Step 3: Solve for \(x\) on Interval [0, \(2\pi)\)

Since sine is opposite at its supplementary angle, \(x\) and \(2x\) need to be supplementary angles and fall in the interval [0, \(2\pi)\). The solution set will be \(x = 2\pi n - \frac{2\pi}{3}\) and \(x = 2\pi n - \frac{4\pi}{3}\) for integer values of \(n\). However, because we are looking for solutions in the interval [0, \(2\pi)\), and we cannot have negative angles in this interval, we only keep the solutions \(\frac{2\pi}{3}\) and \(\frac{4\pi}{3}\).

Key Concepts

Trigonometric Identities

Trigonometric identities are like little mathematical shortcuts. They help us simplify and solve trigonometric equations more easily. In this exercise, we used an important identity:

• The double angle identity: \( 2 \sin{x} \cos{x} = \sin{2x} \)

This identity transforms a product of sine and cosine into a single sine of a double angle, simplifying the equation. Recognizing these identities is crucial because they help us see patterns and simplify expressions quickly, making equations easier to solve.

Often, trigonometric identities are used to relate different trig functions or to break down complex expressions. They are based on the unit circle's properties, allowing equivalent expressions based on angle measures. Remembering and applying these identities correctly can be a real-time saver in solving complex trigonometric equations.

Interval Notation

Interval notation is a concise way of describing a range of values or "interval" on the number line. It's especially useful in mathematics when we're dealing with intervals of solutions, just like in this problem.

For this exercise, the equation needed to be solved on the interval

• \([0, 2\pi)\): This means any solution must be between \(0\) and \(2\pi\), including \(0\) but not including \(2\pi\). The parentheses indicate an open interval on that side.

This notation clearly shows the scope within which we're looking for solutions, which is crucial because trigonometric functions repeat their values in regular intervals (every full rotation in radians).

Using interval notation allows mathematicians and students alike to be very precise and clear about which solutions are valid given a problem's constraints, as it easily distinguishes between different types of intervals and endpoints.

Sine Function

The sine function is one of the basic trigonometric functions that exhibits periodic behavior. Its graph is a smooth, wave-like curve that repeats every \(2\pi\) radians, which is known as its period.

In equations like \(\sin{x} + \sin{2x} = 0\), understanding sine's properties helps solve for the variable \(x\). Here are some key points about the sine function:

• It oscillates between -1 and 1.
• It is positive in the first and second quadrants of the unit circle, and negative in the third and fourth.
• Its zeros (where it crosses the x-axis) are at every integer multiple of \(\pi\).

These properties help in determining solutions within specific intervals. With its predictable pattern, sine makes predicting and finding solutions straightforward, provided we keep within the given range (like the interval \([0, 2\pi)\) in this exercise). Solving the trigonometric equation involved seeing how solutions fitted into sine's recurring pattern, resulting in the interesting answers of \(\frac{2\pi}{3}\) and \(\frac{4\pi}{3}\), where the sine values mirror its waveform behavior.