To solve the equation [tex]\(6 \sin (8x) + 2 = -3\)[/tex], we need to isolate [tex]\(x\)[/tex] step-by-step:
1. Subtract 2 from both sides of the equation:
[tex]\[
6 \sin (8x) = -5
\][/tex]
2. Divide both sides by 6:
[tex]\[
\sin (8x) = -\frac{5}{6}
\][/tex]
3. Find the angle [tex]\( \theta \)[/tex] whose sine is [tex]\(-\frac{5}{6}\)[/tex]:
[tex]\[
8x = \arcsin\left(-\frac{5}{6}\right)
\][/tex]
You can use a calculator or known values to find:
[tex]\[
8x \approx -56.44^\circ
\][/tex]
4. Since sine is periodic, we also have another primary angle:
[tex]\[
8x = 180^\circ - (-56.44^\circ) \approx 236.44^\circ
\][/tex]
5. General solution for sine function periodicity:
The general solution for [tex]\( \sin(\theta) = \sin(k) \)[/tex] is given by:
[tex]\[
8x = -56.44^\circ + n \cdot 360^\circ \quad \text{and} \quad 8x = 236.44^\circ + n \cdot 360^\circ
\][/tex]
6. Dividing everything by 8 to isolate [tex]\( x \)[/tex]:
[tex]\[
x = -56.44^\circ / 8 + n \cdot 360^\circ / 8 \quad \text{and} \quad x = 236.44^\circ / 8 + n \cdot 360^\circ / 8
\][/tex]
Simplifying these expressions:
[tex]\[
x \approx -7.06^\circ + n \cdot 45^\circ \quad \text{and} \quad x \approx 29.56^\circ + n \cdot 45^\circ
\][/tex]
However, examining the original problem, we correctly need:
[tex]\[
x = \theta_1 + n \cdot 45^\circ \quad \text{and} \quad x = \theta_2 + n \cdot 45^\circ
\][/tex]
With:
[tex]\[
\theta_1 = \arcsin\left(-\frac{5}{6}\right) \approx -56.44^\circ / 8 \approx -7.06^\circ
\][/tex]
[tex]\[
\theta_2 = 180^\circ - \arcsin\left(-\frac{5}{6}\right) \approx 236.44^\circ / 8 \approx 29.56^\circ
\][/tex]
To summarize, the solutions that apply to the equation [tex]\(6 \sin (8x)+2=-3\)[/tex] are:
- A [tex]\( -123.56^\circ + n \cdot 360^\circ \)[/tex]
- B [tex]\( -56.44^\circ + n \cdot 360^\circ \)[/tex]
