Question

find the greatest value of 2 sin^2 x - 3 sin x+ 4

Answer 1

Let y = 2sin²x -3sinx + 4

differentiate wrt x
dy/dx = 2sinx.cosx -3cosx

equate to zero ,
dy/dx = 0 = 2sinx.cosx - 3cosx

cosx(2sinx -3) = 0
cosx = 0 and sinx = 3/2
but sinx maximum value 3/2
so, sinx ≠ 3/2

hence,
cosx = 0
then, x = (2n +1)π/2 where n is integer.

now, again, differentiate wrt x
d²y/dx² = 2cos²x -2sin²x +3sinx
= 2(cos²x - sin²x) + 3sinx
= 2cos2x + 3sinx
put x = π/2

d²y/dx² = -2 + 3 = 1 > 0 hence, x = π/2 give maxima not minima ,
so, check again,
x = -π/2

d²y/dx² = -2 -3 < 0
hence, at x = -π/2 function gain, maximum .

so, put x = -π/2 in y

y = 2sin²(-π/2) -3sin(-π/2) + 4
= 2 × 1 -3(-1) + 4
= 2 + 3 + 4
= 9

hence, maximum value of 2sin²x -3sinx +4 = 9