Question

How do I do this question

Answer 1

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)` }

Answer 2

Here is the solution