Question

Solve on the interval [0, 2π):

2sin^2x+3sinx+1=0

A. x= 3π/2, x=7π/6, x=11π/6
B. x=π,x=2π/3, x=5π/3
C. x=2π, x=π/3
D. x=2π, x=π/4, x= 5π/4

Answer 1

Lets x = sinx

\[2x^2 + 3x +1 = 0\]
\[(2x +1)(x +1)=0\]
\[x = \frac{-1}{2} , -1\]

Answer 2

\[sinx = \frac{-1}{2},-1\]
when x = 3pi/2 ,sinx =-1
So A

Answer 3

option A

Answer 4

did u understand @kassia