Questions: How many solutions are there to the following equation over the interval [0,2 π) ? 2 sin ^2(3 x)+2 sin (3 x)=0 A) 3 B) 4 C) 6 D) 9 E) 12 F) None of the above

Solution Steps

To solve the equation \(2 \sin^2(3x) + 2 \sin(3x) = 0\) over the interval \([0, 2\pi)\), we first factor the equation. This gives us a product of terms that we can set to zero to find the solutions for \(\sin(3x)\). We then solve for \(x\) by considering the periodic nature of the sine function and the given interval.

We start with the equation: \[ 2 \sin^2(3x) + 2 \sin(3x) = 0 \] Factoring out \(2 \sin(3x)\), we get: \[ 2 \sin(3x)(\sin(3x) + 1) = 0 \]

Setting each factor to zero gives us two equations to solve:

1. \( \sin(3x) = 0 \)
2. \( \sin(3x) = -1 \)

The general solutions for \( \sin(3x) = 0 \) are: \[ 3x = n\pi \quad \text{for } n \in \mathbb{Z} \] Thus, we have: \[ x = \frac{n\pi}{3} \] For \( n = 0, 1, 2, 3, 4, 5 \), the valid solutions in the interval \([0, 2\pi)\) are: \[ x = 0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3} \]

The general solution for \( \sin(3x) = -1 \) is: \[ 3x = \frac{3\pi}{2} + 2k\pi \quad \text{for } k \in \mathbb{Z} \] Thus, we have: \[ x = \frac{\frac{3\pi}{2} + 2k\pi}{3} = \frac{\pi}{2} + \frac{2k\pi}{3} \] For \( k = 0, 1 \), the valid solutions in the interval \([0, 2\pi)\) are: \[ x = \frac{\pi}{2}, \frac{7\pi}{6} \]

Combining all valid solutions from both cases, we have:

• From \( \sin(3x) = 0 \): \( 0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3} \) (6 solutions)
• From \( \sin(3x) = -1 \): \( \frac{\pi}{2}, \frac{7\pi}{6} \) (2 solutions)

In total, we have \(6 + 2 = 8\) solutions.

Final Answer