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

Question

Anna Andrews · Accepted Answer

Solve 8sin^2 (3x) = - 7sin (3x)

Ans: #-20^@35 and 80^@35# #8sin^2 (3x) + 7sin (3x) = 0# Put sin (3x) in common factor: #sin (3x)(8sin 3x + 7) = 0# a. #sin 3x = 0# --> 3x = 0 --> x = 0 #sin 3x = pi# --> #x = pi/3# #sin 3x = 2pi# --> #x = (2pi)/3# b. 8sin (3x) + 7 = 0 --> #sin 3x = -7/8# 3x = - 61.04 --> x = #-20^@35# 3x = 180 - (-61.04) = 241.04 --> #x = 80^@35# Check by calculator x = 80.35 3x = 241.04 --> sin 3x = -0.875 --> #sin^2 (3x) = 0.77# #8sin^2 (3x) = 0.77(8) = 6.12# --> #-7sin 3x = 7(0.875) = 6.125#. OK

Violet Anderson · Answer

undefined 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.