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