brianpathmasri brianpathmasri

12-03-2021
Mathematics

contestada

sin^2 x + 3sin x + 2 = 0f for 0 <= x <= 2pi

Solve for x

Worth 8 marks.

Respuesta :

Shrabonisau Shrabonisau

12-03-2021

Step-by-step explanation:

first inequality gives

(2sinx−1)(sinx+2)>0

(sinx−

2

1

)>0

6

Π

<x<

6

5Π

second inequality gives

(x+1)(x−2)<0

−1<x<2

common part is

6

Π

<x<2

hOpe it help!!!

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