Answer:
[tex]S = S_{1} \cup S_{2}[/tex], [tex]S_{1} = \left\{0, \frac{\pi}{2}, \pi,\frac{3\pi}{2}, 2\pi \right\}[/tex] and [tex]S_{2} = \left\{\frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6} \right\}[/tex]
Step-by-step explanation:
The equation needs to rearranged in terms of solely one trigonometric function if possible:
[tex]\sin 2x - 2\cdot \sin 2x \cdot \cos 2x = 0[/tex]
[tex]\sin 2x \cdot (1 - 2\cdot \cos 2x) = 0[/tex]
Which means that expression is equal to zero if [tex]\sin 2x = 0[/tex] or [tex]1 - 2\cdot \cos 2x = 0[/tex]
Case I - [tex]\sin 2x = 0[/tex]
[tex]x = \frac{1}{2}\cdot \sin^{-1} 0[/tex]
[tex]S_{1} = \left\{0, \frac{\pi}{2}, \pi,\frac{3\pi}{2}, 2\pi \right\}[/tex]
Case II - [tex]1 - 2\cdot \cos 2x = 0[/tex]
[tex]2\cdot \cos 2x = 1[/tex]
[tex]\cos 2x = \frac{1}{2}[/tex]
[tex]2x = \cos^{-1} \frac{1}{2}[/tex]
[tex]x = \frac{1}{2}\cdot \cos^{-1} \frac{1}{2}[/tex]
[tex]S_{2} = \left\{\frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6} \right\}[/tex]
The complete set of solution is:
[tex]S = S_{1} \cup S_{2}[/tex]
