How do I do this question

Question

asked Jul 26, 2023 60.3k views

2 Answers

MBender · Answer 1 · 2023-07-28T14:44:40+0000

Answer:

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 ,
$\pi$ ,
$2\pi$

2 cos x = - 1

cos x =
-(1)/(2) ⇒ x =
$(2\pi )/(3)$ ,
$(4\pi )/(3)$

x ∈ { 0 ,
$(2\pi )/(3)$ ,
$\pi$ ,
$(4\pi )/(3)$ ,
$2\pi$ }

(b). sin 2x - 2 cos x = 0 ( 0 ≤ x ≤ 2π )

2 cos x ( sin x - 1 ) = 0

cos x = 0 ⇒ x =
$(\pi )/(2)$ ,
$(3\pi )/(2)$

sin x = 1 ⇒ x =
$(\pi )/(2)$ ,
$(3\pi )/(2)$

x ∈ {
$(\pi )/(2)$ ,
$(3\pi )/(2)$ }