Answer:
Step-by-step explanation:
(a). sin2x + sinx = 0 ( 0 ≤ x ≤ 2π )
sin2x = 2sinx(cosx)
2 ( sin x )( cos x ) + sin x = 0
sin x ( 2 cos x + 1 ) = 0
sin x = 0 ⇒ x = 0 , [tex]\pi[/tex] , [tex]2\pi[/tex]
2 cos x = - 1
cos x = [tex]-\frac{1}{2}[/tex] ⇒ x = [tex]\frac{2\pi }{3}[/tex] , [tex]\frac{4\pi }{3}[/tex]
x ∈ { 0 , [tex]\frac{2\pi }{3}[/tex] , [tex]\pi[/tex] , [tex]\frac{4\pi }{3}[/tex] , [tex]2\pi[/tex] }
(b). sin 2x - 2 cos x = 0 ( 0 ≤ x ≤ 2π )
2 cos x ( sin x - 1 ) = 0
cos x = 0 ⇒ x = [tex]\frac{\pi }{2}[/tex] , [tex]\frac{3\pi }{2}[/tex]
sin x = 1 ⇒ x = [tex]\frac{\pi }{2}[/tex] , [tex]\frac{3\pi }{2}[/tex]
x ∈ { [tex]\frac{\pi }{2}[/tex] , [tex]\frac{3\pi }{2}[/tex] }
