Solve sin2∅=sin∅ on the interval 0≤x< 2\pi . a. 0,(\pi )/(3) b. 0, \pi, (\pi )/(3), (5\pi )/(3) c. 0, \pi, (2\pi )/(3),(4\pi )/(3) d. (3\pi )/(2), (\pi )/(2),(\pi )/(6), (5\p…

Question

Solve sin2∅=sin∅ on the interval 0≤x< 2
`\pi` .

a. 0,
`(\pi )/(3)`
b. 0,
`\pi`,
`(\pi )/(3)`,
`(5\pi )/(3)`
c. 0,
`\pi`,
`(2\pi )/(3)`,
`(4\pi )/(3)`
d.
`(3\pi )/(2)`,
`(\pi )/(2)`,
`(\pi )/(6)`,
`(5\pi )/(6)`

Answer 1

`\large\boxed{b.\ 0,\ \pi,\ (\pi)/(3),\ (5\pi)/(3)}`

Explanation:

`\text{Use}\ \sin2x=2\sin x\cos x\\\\\sin2\O=\sin\O\\\\2\sin\O\cos\O=\sin\O\qquad\text{subtract}\ \sin\O\ \text{from both sides}\\\\2\sin\O\cos\O-\sin\O=0\qquad\text{distribute}\\\\\sin\O(2\cos\O-1)=0\iff\sin\O=0\ \vee\ 2\cos\O-1=0`

`\sin\O=0\iff\O=0\ \vee\ \O=\pi\\\\2\cos\O-1=0\qquad\text{add 1 to both sides}\\\\2\cos\O=1\qquad\text{divide both sides by 2}\\\\\cos\O=(1)/(2)\iff\O=(\pi)/(3)\ \vee\ \O=(5\pi)/(3)`