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Solve sin2∅=sin∅ on the interval 0≤x< 2[tex]\pi[/tex] .
a. 0,[tex]\frac{\pi }{3}[/tex]
b. 0, [tex]\pi[/tex], [tex]\frac{\pi }{3}[/tex], [tex]\frac{5\pi }{3}[/tex]
c. 0, [tex]\pi[/tex], [tex]\frac{2\pi }{3}[/tex],[tex]\frac{4\pi }{3}[/tex]
d. [tex]\frac{3\pi }{2}[/tex], [tex]\frac{\pi }{2}[/tex],[tex]\frac{\pi }{6}[/tex], [tex]\frac{5\pi }{6}[/tex]

Respuesta :

gmany

Answer:

[tex]\large\boxed{b.\ 0,\ \pi,\ \dfrac{\pi}{3},\ \dfrac{5\pi}{3}}[/tex]

Step-by-step explanation:

[tex]\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[/tex]

[tex]\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=\dfrac{1}{2}\iff\O=\dfrac{\pi}{3}\ \vee\ \O=\dfrac{5\pi}{3}[/tex]

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