Question

1. sin6x/cos8x=-1 2. sin12x/sin8x=-1

Answer 1

$1. \ \frac{sin6x}{cos8x} = -1\\\\ cos8x \ne 0; \ 8x \ne \frac{\pi}{2} + \pi n,\ n \in Z; \ x \ne \frac{\pi}{16} + \frac{\pi n}{8}, \ n \in Z\\\\ sin6x = -cos8x\\\\ sin6x + cos8x = 0\\\\ cos(\frac{\pi}{2} - 6x) + cos8x = 0\\\\ 2cos(\frac{\frac{\pi}{2} - 6x + 8x}{2})cos(\frac{\frac{\pi}{2} - 6x - 8x }{2}) = 0\\\\ cos(\frac{\pi}{4} + x)cos(\frac{\pi}{4} - 7x) = 0$

$1) \ cos(\frac{\pi}{4} + x) = 0\\\\ \frac{\pi}{4} + x = \frac{\pi}{2} + \pi n, \ n \in Z\\\\ x = \frac{\pi}{4} + \pi n, \ n \in Z \\\\ \frac{\pi}{4} + \pi n \ne \frac{\pi}{16} + \frac{\pi k}{8}\\\\ \ n \ne \frac{2k - 3}{16}\\\\ \boxed{ x = \frac{\pi}{4} + \pi n, \ n \in Z \setminus \{ \frac{2k - 3}{16}| \ k \in Z \} }$

$2) \ cos(\frac{\pi}{4} - 7x) = 0\\\\ \frac{\pi}{4} - 7x =\frac{\pi}{2} + \pi n, \ n \in Z\\\\ -7x = \pi n + \frac{\pi}{4}, \ n \in Z\\\\ -\frac{\pi n}{7} - \frac{\pi}{28} \ne \frac{\pi}{16} + \frac{\pi k}{8}\\\\ \ n \ne -\frac{14k + 11}{16}\\\\ \boxed{ x = -\frac{\pi n}{7} - \frac{\pi}{28}, \ n \in Z \setminus \{ -\frac{14k + 11}{16}| \ k \in Z \} }$

$2. \ \frac{sin12x}{sin8x} = -1\\\\ sin8x \ne 0; \ 8x \ne \pi n, \ n \in Z; \ x \ne \frac{\pi n}{8}, \ n \in Z\\\\ sin12x + sin8x = 0\\\\ 2sin(\frac{12x + 8x}{2})cos(\frac{12x-8x}{2}) = 0\\\\ sin(10x)cos(4x) = 0\\\\\\ 1) \ sin(10x) = 0\\\\ 10x = \pi n, \ n \in Z,\\\\ \frac{\pi n}{10} \ne \frac{\pi k}{8}\\\\ n \ne \frac{5k}{4} \\\\ \boxed{ x = \frac{\pi n}{10}, \ n \in Z \setminus \{ \frac{5k}{4} | \ k \in Z \} }$

$2) \ cos(4x) = 0\\\\ 4x = \frac{\pi}{2} + \pi n, \ n \in Z\\\\ x = \frac{\pi}{8} + \frac{\pi n}{4}, \ n \in Z\\\\ \frac{\pi}{8} + \frac{\pi n}{4} \ne \frac{\pi k}{8} \\\\ n \ne \frac{k - 1}{2}\\\\ \boxed{ x = \frac{\pi}{8} + \frac{\pi n}{4}, \ n \in Z \setminus \{\frac{k - 1}{2} | \ k \in Z \} }$