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

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

Ans:

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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:

- (y = 0)
- (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.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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