3 years ago

Sin^2 x + 3sin x + 2 = 0f for 0 <= x <= 2pi Solve for x Worth 8 marks.

Mathematics

1 answer:

makvit [3.9K]3 years ago

5 0

Step-by-step explanation:

first inequality gives

(2sinx−1)(sinx+2)>0

(sinx−

)>0

<x<

5Π

second inequality gives

(x+1)(x−2)<0

−1<x<2

common part is

<x<2

hOpe it help!!!

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