( 0 , \frac { \pi } { 6 } ) \cup ( \frac { 5 \pi } { 6 } , 2 \pi ) ( b

1.	( 0 , \frac { \pi } { 6 } ) \cup ( \frac { 5 \pi } { 6 } , 2 \pi ) ( b ) ( 0 , \frac { 5 \pi } { 6 } ) \cup ( \pi , 2 \pi ) \quad ( c ) ( 0 , \frac { \pi } { 6 } ) \cup ( \pi , 2 \pi )
Answer» Given the equation is. 2sin²∅-5sin∅+2 > 0=> 2sin²∅-4sin∅-sin∅+2 > 0=> 2sin∅(sin∅-2)-1(sin∅-2)>0=> (sin∅-2)(2sin∅-1) > 0=> sin∅ = [-1,1/2) but sin∅ has range from [-1,1] so, sin∅ belongs to. (0,π/6) U (5π/6,2π) nice solution

( 0 , \frac { \pi } { 6 } ) \cup ( \frac { 5 \pi } { 6 } , 2 \pi ) ( b ) ( 0 , \frac { 5 \pi } { 6 } ) \cup ( \pi , 2 \pi ) \quad ( c ) ( 0 , \frac { \pi } { 6 } ) \cup ( \pi , 2 \pi )

Answer»

Given the equation is.

2sin²∅-5sin∅+2 > 0=> 2sin²∅-4sin∅-sin∅+2 > 0=> 2sin∅(sin∅-2)-1(sin∅-2)>0=> (sin∅-2)(2sin∅-1) > 0=> sin∅ = [-1,1/2)

but sin∅ has range from [-1,1]

so, sin∅ belongs to. (0,π/6) U (5π/6,2π)

nice solution