Question

Solve on the interval [0,2π):
(sinx+1)(2sin^2 x-3sinx - 2) = 0

Answer 1

\[(a+1)(2a^2-3a-2)=0\]
\[a=-1\]or\[ 2a^2-3a-2=0\]
\[(2a+1)(a-2)\]
\[a=-1/2,a=2\]

Answer 2

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

Answer 3

put a=sin x

Answer 4

From my work, it's D. Am I wrong?