How do you find the solutions to #8sin^2(3x) = -7sin(3x)#?

To find the solutions to the equation $8 \sin^2(3x) = -7 \sin(3x)$, first, rewrite it as a quadratic equation in terms of $\sin(3x)$:

$8 \sin^2(3x) + 7 \sin(3x) = 0$

Let $y = \sin(3x)$. Then the equation becomes:

$8y^2 + 7y = 0$

Now, factor out $y$:

$y(8y + 7) = 0$

This equation has two solutions:

1. $y = 0$
2. $8y + 7 = 0$

For $y = 0$, it implies:

$\sin(3x) = 0$

Solving this equation, we get:

$3x = n\pi, \text{ where } n \text{ is an integer}$

For $8y + 7 = 0$, it implies:

$8y = -7$

Solving for $y$, we get:

$y = -\frac{7}{8}$

Substituting back $\sin(3x) = -\frac{7}{8}$, we get:

$3x = \arcsin\left(-\frac{7}{8}\right) + 2n\pi \text{ or } 3x = \pi - \arcsin\left(-\frac{7}{8}\right) + 2n\pi$

where $n$ is an integer.

So, the solutions to the equation $8 \sin^2(3x) = -7 \sin(3x)$ are:

$x = \frac{\arcsin\left(-\frac{7}{8}\right)}{3} + \frac{2n\pi}{3} \text{ or } x = \frac{\pi - \arcsin\left(-\frac{7}{8}\right)}{3} + \frac{2n\pi}{3}$

where $n$ is an integer.