65.0k views
5 votes
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)

User Alex Shkor
by
5.2k points

1 Answer

5 votes

Answer:


\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)

User OrdoDei
by
4.9k points