Question

Find all solutions to the equation in the interval [0, 2π). sin 2x - sin 4x = 0

Answer 1

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:

$S = S_{1} \cup S_{2}$

Answer 2

Sin 2x - sin 4x = 0
sin 2x - 2sin 2x cos 2x = 0
sin x(1 - 2cos 2x) = 0
sin x = 0 or 1 - 2cos 2x = 0
sin x = 0 or 2cos 2x = 1
sin x = 0 or cos 2x = 1/2
x = arc sin 0 or 2x = arc cos (1/2)
x = arc sin 0 or x = 1/2 arc cos (1/2)
x = 0, π, 2π or x = π/6, 7π/6