+318 Solve the equation
2sin ^2 θ-3 sin θ+1=0
What is the solution set in the interval 0
+23252 2sin2(x) - 3sin(x) + 1 = 0
( 2sin(x) - 1 )( sin(x) - 1 ) = 0
Either: 2sin(x) - 1 = 0 ---> 2sin(x) = 1 ---> sin(x) = 1/2 ---> x = sin-1( 1/2 )
---> x = pi/6 or x = 5pi/6
Or: sin(x) - 1 = 0 ---> sin(x) = 1 ---> x = sin-1(1) ---> x = pi/2
[ plus or minus all multiples of 2pi ]
