Question

4sincubx-6sinsquarex+12sinx+100

Answer 1

f(x) = 4sin^3x -6sin^2x+12sinx +100
any function decreasing when
f'( x) < 0
now differentiate w.r.t x

f '( x) =12sin^2x.cosx -12sinx.cosx +12cosx
for decreasing
f'(x ) < 0
12cosx{ sin^2x-sinx + 1} < 0
this is possible
case 1 :- cosx < 0 and sin^2x -sinx +1 >0
if you see graph of cosx
cosx < 0 in ( -π/2 , π/2 ) , (π, 3π/2 ) etc

now,
sin^2x -sinx +1 > 0
this is always right
hence , x € R
now, common value of both solution of x is ( -π/2 , π/2) , (π/2 , 3π/2 ) ..
e.g x € { (2n-1)π/2 , (2n+1)π/2 }

case 2 :- when cosx > 0 and
sin^2x -sinx +1 <0
but sin^2 -sinx +1 is always positive
so, case 2 has no any solution ,

hence,
x € {(2n-1)π/2 , (2n+1)π/2 } only solution .