a) For the equation \(\sin \left(5x - \frac{\pi}{3}\right) = -1\), we need to find the values of \(5x - \frac{\pi}{3}\) that make the sine function equal to -1. The sine function equals -1 at \(\frac{3\pi}{2} + 2k\pi\) for any integer \(k\). We then solve for \(x\) within the interval \([0, 2\pi)\).
b) For the equation \(\sin^2(4\alpha) + 5\sin(4\alpha) - 6 = 0\), we treat \(\sin(4\alpha)\) as a variable, say \(y\). This transforms the equation into a quadratic equation \(y^2 + 5y - 6 = 0\). We solve for \(y\) and then find the corresponding \(\alpha\) values within the interval \([0, 2\pi)\).
We need to solve the equation
\[
\sin \left(5x - \frac{\pi}{3}\right) = -1
\]
The sine function equals \(-1\) at
\[
5x - \frac{\pi}{3} = \frac{3\pi}{2} + 2k\pi
\]
for any integer \(k\). Rearranging gives:
\[
5x = \frac{3\pi}{2} + \frac{\pi}{3} + 2k\pi
\]
Calculating the common angle:
\[
\frac{3\pi}{2} + \frac{\pi}{3} = \frac{9\pi}{6} + \frac{2\pi}{6} = \frac{11\pi}{6}
\]
Thus,
\[
5x = \frac{11\pi}{6} + 2k\pi
\]
Dividing by 5:
\[
x = \frac{11\pi}{30} + \frac{2k\pi}{5}
\]
Evaluating for \(k = 0, 1, 2, 3, 4\) gives the solutions:
- For \(k = 0\): \(x \approx 1.152\)
- For \(k = 1\): \(x \approx 2.409\)
- For \(k = 2\): \(x \approx 3.665\)
- For \(k = 3\): \(x \approx 4.922\)
- For \(k = 4\): \(x \approx 6.178\)
We need to solve the equation
\[
\sin^2(4\alpha) + 5\sin(4\alpha) - 6 = 0
\]
Letting \(y = \sin(4\alpha)\), we rewrite the equation as:
\[
y^2 + 5y - 6 = 0
\]
Using the quadratic formula:
\[
y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{25 + 24}}{2} = \frac{-5 \pm 7}{2}
\]
This gives us the roots:
\[
y_1 = 1 \quad \text{and} \quad y_2 = -6
\]
Since \(-1 \leq y \leq 1\), we only consider \(y_1 = 1\). Thus,
\[
\sin(4\alpha) = 1
\]
This occurs at
\[
4\alpha = \frac{\pi}{2} + 2k\pi
\]
Solving for \(\alpha\):
\[
\alpha = \frac{\pi}{8} + \frac{k\pi}{2}
\]
Evaluating for \(k = 0, 1, 2, 3\) gives the solutions:
- For \(k = 0\): \(\alpha \approx 0.393\)
- For \(k = 1\): \(\alpha \approx 1.963\)
- For \(k = 2\): \(\alpha \approx 3.534\)
- For \(k = 3\): \(\alpha \approx 5.105\)
The solutions for part a are:
\[
\boxed{x \approx 1.152, 2.409, 3.665, 4.922, 6.178}
\]
The solutions for part b are:
\[
\boxed{\alpha \approx 0.393, 1.963, 3.534, 5.105}
\]
Answered
