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
Since, $$2sin2\theta=1$$ and $$sin2\theta=\frac{1}{2}$$
Thus, $$\theta=\frac{\pi}{12},\frac{5\pi}{12}$$
when $$2sin2\theta=-1$$
Thus, $$sin2\theta=-\frac{1}{2}$$
So, $$\theta=\frac{7\pi}{12},\frac{11\pi}{12}$$
Point of intersections will be :
$$(1,\frac{\pi}{12}), (1,\frac{5\pi}{12}),(1,\frac{13\pi}{12}),(1,\frac{17\pi}{12}),(-1,\frac{7\pi}{12}),(-1,\frac{11\pi}{12}),(-1,\frac{19\pi}{12}),(-1,\frac{23\pi}{12})$$
