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