Question

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

Answer 1

`S = S_(1) \cup S_(2)`,
`S_(1) = \left\{0, (\pi)/(2), \pi,(3\pi)/(2), 2\pi \right\}` and
`S_(2) = \left\{(\pi)/(6), (5\pi)/(6), (7\pi)/(6), (11\pi)/(6) \right\}`

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 = (1)/(2)\cdot \sin^(-1) 0`

`S_(1) = \left\{0, (\pi)/(2), \pi,(3\pi)/(2), 2\pi \right\}`

Case II -
`1 - 2\cdot \cos 2x = 0`

`2\cdot \cos 2x = 1`

`\cos 2x = (1)/(2)`

`2x = \cos^(-1) (1)/(2)`

`x = (1)/(2)\cdot \cos^(-1) (1)/(2)`

`S_(2) = \left\{(\pi)/(6), (5\pi)/(6), (7\pi)/(6), (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