another trigo equ general solution
p213 q11
Your answers and the books answers are equivalent, but I much prefer your answers! Observe:
Book answers:
$\displaystyle \theta = \pm \frac{\pi}{2} + 2n\pi = \pm \frac{\pi}{2}, \, \frac{3\pi}{2}, \frac{5\pi}{2}, \, ....$
$\displaystyle \theta = \frac{n \pi}{3} = 0 \, \pm \frac{\pi}{3}, \, \pm \frac{2\pi}{3}, \pm \frac{3\pi}{3} = \pm \pi, \, ....$
These answers can be re-grouped as:
$\displaystyle \theta = 0, \, \pm \frac{\pi}{2}, \, \pm \pi, \, \pm \frac{3\pi}{2}, \, ......$
$\displaystyle \theta = \pm \frac{\pi}{3}, \, \frac{2 \pi}{3}, \, \frac{4 \pi}{3}, \, \frac{5 \pi}{3}, \, \frac{7 \pi}{3}, \, .....$
The first group is given in general by $\displaystyle \theta = \frac{n \pi}{2}$ and the second group is given by $\displaystyle \theta = \pm \frac{\pi}{3} + n \pi$, where n is an integer. These are your answers.
Which is best? It doesn't matter! (But I like yours).
With regards to method 2, you use the fact that $\displaystyle \sin A = \sin B \Rightarrow A = B + 2n\pi$ or $\displaystyle A = (\pi - B) + 2 n \pi$.
So $\displaystyle \sin(4 \theta) = \sin(-2 \theta)\, $ yields:
$\displaystyle 4 \theta = -2 \theta + 2n\pi \Rightarrow 6 \theta = 2n\pi \Rightarrow \theta = \frac{n \pi}{3}$, one of the answers given in the book,
or
$\displaystyle 4 \theta = (\pi - [-2 \theta]) + 2n\pi = \pi + 2 \theta + 2n \pi \Rightarrow 2 \theta = \pi + 2n\pi \Rightarrow \theta = \frac{\pi}{2} + n \pi$. You can confirm that this is equivalent to $\displaystyle \theta = \pm \frac{\pi}{2} + 2n \pi$, the other answer given in the book.
Method 2 looks like the method the book has used. But I don't know why they decided to use the general solution $\displaystyle \theta = \pm \frac{\pi}{2} + 2n \pi$ rather than $\displaystyle \theta = \frac{\pi}{2} + n \pi$ ....