Answer
$(1,\frac{\pi}{12}), (1,\frac{5\pi}{12}),(1,\frac{13\pi}{12}),(1,\frac{17\pi}{12}),(-1,\frac{7\pi}{12}),(-1,\frac{11\pi}{12}),(-1,\frac{19\pi}{12}),(-1,\frac{23\pi}{12})$
Work Step by Step
$2sin2\theta=1$
$sin2\theta=\frac{1}{2}$
$\theta=\frac{\pi}{12},\frac{5\pi}{12}$
Now,
$2sin2\theta=-1$
$sin2\theta=\frac{1}{2}$
$\theta=\frac{7\pi}{12},\frac{11\pi}{12}$
Therefore, the point of intersections are:
$(1,\frac{\pi}{12}), (1,\frac{5\pi}{12}),(1,\frac{13\pi}{12}),(1,\frac{17\pi}{12}),(-1,\frac{7\pi}{12}),(-1,\frac{11\pi}{12}),(-1,\frac{19\pi}{12}),(-1,\frac{23\pi}{12})$
