`2sin^2x+3sinx-2>0 and x^2-x-2<0.` Find the solution set.

Let 2sin^2x+3sinx-2>0 andx^2-x-2<0(x is measured in radians). Then x lies in the interval (a) (pi/6,(5pi)/6) (b) (-1,(5pi)/6) (c) (-1,2) (d) (pi/6,2)

2sin^2x-7sinx +3=0

lim_(x to pi//6) (2sin^2x+sinx-1)/(2sin^2x-3sinx+1)=

Evaluate the following limits: lim_(x to(pi)/(6))(2sin^(2)x+sinx-1)/(2sin^(2)x-3sinx+1)

lim_(xto(pi)/(6))(2sin^(2)x+sinx-1)/(2sin^(2)x-3sinx+1)=

lim_(x rarr pi/6) (2sin^(2)x + sin x-1)/(2sin^(2)x-3sinx -1) =