Question

The sum of all values of \(x\) in \([0, 2\pi]\), for which \(x + \sin(2x) + \sin(3x) + \sin(4x) = 0\) is equal to:

• A. \(8\pi\)
• B. \(11\pi\)
• C. \(12\pi\)
• D. \(9\pi\)

Hint:

Check solutions by substituting back into the original equation to confirm their validity.

Answer 1

The Correct Option is D

We are given the equation: \[ \left( \sin x + \sin 4x \right) + \left( \sin 2x + \sin 3x \right) = 0 \] We begin by simplifying this equation using trigonometric identities. Using the sum-to-product identity: \[ \sin A + \sin B = 2 \sin \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right) \] We can apply this identity to both pairs of sines: \[ \left( \sin x + \sin 4x \right) = 2 \sin \left( \frac{x + 4x}{2} \right) \cos \left( \frac{4x - x}{2} \right) = 2 \sin \left( \frac{5x}{2} \right) \cos \left( \frac{3x}{2} \right) \] \[ \left( \sin 2x + \sin 3x \right) = 2 \sin \left( \frac{2x + 3x}{2} \right) \cos \left( \frac{3x - 2x}{2} \right) = 2 \sin \left( \frac{5x}{2} \right) \cos \left( \frac{x}{2} \right) \] Thus, the equation becomes: \[ 2 \sin \left( \frac{5x}{2} \right) \cos \left( \frac{3x}{2} \right) + 2 \sin \left( \frac{5x}{2} \right) \cos \left( \frac{x}{2} \right) = 0 \] Factor out \( 2 \sin \left( \frac{5x}{2} \right) \): \[ 2 \sin \left( \frac{5x}{2} \right) \left( \cos \left( \frac{3x}{2} \right) + \cos \left( \frac{x}{2} \right) \right) = 0 \] For this product to be zero, either: \[ \sin \left( \frac{5x}{2} \right) = 0 \quad {or} \quad \cos \left( \frac{3x}{2} \right) + \cos \left( \frac{x}{2} \right) = 0 \] Let's first solve for the case where \( \sin \left( \frac{5x}{2} \right) = 0 \): \[ \sin \left( \frac{5x}{2} \right) = 0 \quad \Rightarrow \quad \frac{5x}{2} = 0, \pi, 2\pi, 3\pi, \dots \] Thus, \( x = 0, \frac{2\pi}{5}, \frac{4\pi}{5}, \frac{6\pi}{5}, \dots \). Now, let's solve for the case where \( \cos \left( \frac{3x}{2} \right) + \cos \left( \frac{x}{2} \right) = 0 \): Using the sum-to-product identity again: \[ \cos A + \cos B = 2 \cos \left( \frac{A + B}{2} \right) \cos \left( \frac{A - B}{2} \right) \] We apply this identity to the equation: \[ \cos \left( \frac{3x}{2} \right) + \cos \left( \frac{x}{2} \right) = 2 \cos \left( \frac{3x}{2} + \frac{x}{2} \right) \cos \left( \frac{3x}{2} - \frac{x}{2} \right) \] \[ = 2 \cos \left( \frac{4x}{2} \right) \cos \left( \frac{2x}{2} \right) = 2 \cos (2x) \cos (x) \] Thus, the equation becomes: \[ 2 \cos (2x) \cos (x) = 0 \] For this product to be zero, either: \[ \cos (2x) = 0 \quad {or} \quad \cos (x) = 0 \] For \( \cos (2x) = 0 \): \[ 2x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots \quad \Rightarrow \quad x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \dots \] For \( \cos (x) = 0 \): \[ x = \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \dots \] Now, combining all the solutions, we have the general solution for \( x \): \[ x = 0, \pi, 2\pi, 3\pi, \dots \] Finally, we can conclude that the sum of these values of \( x \) is: \[ 6\pi + \pi + 2\pi = 9\pi \]