Problem 93 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the solutions of the equation in the interval \([\mathbf{0}, \mathbf{2} \pi)\) Use a graphing utility to verify your answers. $$\sin 6 x+\sin 2 x=0$$

Step-by-Step Solution

Answer

The solutions of the equation \( \sin 6x + \sin 2x = 0 \) over the interval \( [0, 2\pi) \) are \( x = 0, π/4, π/2, 3π/4, π, 5π/4, 3π/2, 7π/4, π/2 and 3π/2 \)

Step 1: Apply sine addition formula

First, apply the sine addition formula \( \sin \alpha + \sin \beta = 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} \). Replace \(\alpha\) with \(6x\) and \(\beta\) with \(2x\) to get \( \sin 6x + \sin 2x = 2 \sin4x\cos2x \). Now the equation becomes \(2 \sin 4x \cos 2x = 0 \)

Step 2: Solve for x

Now we need to solve for \(x\). We know in a product 'A x B = 0', if either A=0 or B=0, then the product will be zero. Here, either \( \sin 4x = 0 \) or \( \cos 2x = 0 \)\n\nWhen \( \sin 4x = 0 \), the solution is \(x = k \pi /4\) where \(k\) is an integer\nWhen \( \cos 2x = 0 \), the solution is \(x = (2k+1) \pi /4\) where \(k\) is an integer

Step 3: Apply the Given Interval (0,2π)

The given interval for solution is \([0, 2\pi)\). For this interval, when \(x = k \pi /4\) the solution includes \(x\) in \{0, π/4, π/2, 3π/4, π, 5π/4, 3π/2, 7π/4\}\nWhen \( x = (2k+1) \pi /4 \) the solution includes \(x\) in \{π/2, 3π/2\}

Step 4: Graphing the function

By using a graphing utility, draw the graph of the function \( \sin 6x + \sin 2x \). The x-intercepts of this graph, which are the solutions to the equation, will verify the above calculated solutions.

Key Concepts

Sine Addition Formula

The sine addition formula plays a pivotal role in solving trigonometric equations, especially when dealing with expressions involving the sum of sine functions with different arguments. It is given by \[ \sin\alpha + \sin\beta = 2\sin\frac{\alpha + \beta}{2}\cos\frac{\alpha - \beta}{2} \.\]
When faced with the equation \( \sin 6x + \sin 2x = 0 \) within a textbook, applying this formula is essential. By setting \( \alpha = 6x \) and \( \beta = 2x \) and substituting into the sine addition formula, we simplify the complex-looking equation into a more manageable form involving the products of sine and cosine terms.
The main advantage here is that it turns a sum of sine functions into a product of a sine and cosine function, which we can easily zero out to find solutions since if any factor in a product equals zero, the entire product is zero. This concept is the stepping stone to grasping the method to solve a wide variety of trigonometric equations.

Graphing Utility Verification

Verifying solutions with a graphing utility is essential, as it provides a visual confirmation of the answers and helps understand the behavior of trigonometric functions. After simplifying the given equation \( \sin 6x + \sin 2x = 0 \) to \( 2\sin4x\cos2x = 0 \) and finding the theoretical solutions, using a graphing utility allows us to plot the original equation.
In the plotted graph, the x-intercepts represent the points where the function equals zero, thus visually verifying the solutions obtained algebraically. This method serves not only as a means to check our work but also deepens our understanding of trigonometric functions and their intersections with the x-axis, which correspond to our solutions. Adequate proficiency in using graphing utilities is a must-have skill for students and can greatly aid in interpreting and solving trigonometric equations.

Trigonometric Equations Interval Solutions

Solving for interval solutions in trigonometric equations is about finding all the possible angles within a specific range that satisfy the equation. Through analysis of the equation \( 2\sin4x\cos2x = 0 \) and using the zero product property, we've found that \( x = k\pi/4 \) and \( x = (2k+1)\pi/4 \) where \( k \) is an integer.
However, we're confined to solutions within the interval \( [0, 2\pi) \) which means we must now identify which multiple of \( k \) fits within this range. By listing all possible values of \( x \) that fall within our interval, we determine the complete set of interval solutions. This step is crucial in understanding that trigonometric equations have an infinite number of solutions in their general form, but when we restrict them to a certain interval, we filter out the specific set of relevant solutions.