If 4 sinx+3=0 on `0lexlt2pi`, Then x=

If 4 sin^(2)x-8 sinx+3 le 0,0 le x le 2pi, x in

1+sinx + sin^2(x) + .... = 4+2sqrt3, 0ltxltpi, x!=pi, then x = 1) (pi)/(3), (pi)/(4) 2) (pi)/(4), (pi)/(6) 3) (2 pi)/(5), (pi)/(6) 4) (pi)/(3), (2 pi)/(3)

Find the number of solution(s) of the following questions (a) x^2 + x + 1 + sinx = 0, x in [0, 2pi] (b) x^2 + x + 1 + sinx = 0, x in [pi, 2pi] (c) x^2 + x + sinx = 0, x in [0, pi]

Statement 1: lim x ->0 ((sqrt(1-cos2x)) logcosx)/x^2 does not exist. Statement 2: |sinx| = { sinx, 0

The number of solutions of the equation x^(3)+x^(2)+4x+2sinx=0 in 0 le x le 2pi is

The number of solutions of the equation x^(3)+x^(2)+4x+2sinx=0 in 0 le x le 2pi is

The number of solutions of the equation x^(3)+x^(2)+4x+2sinx=0 in 0 le x le 2pi is

The number of solutions of the equation x^(3)+x^(2)+4x+2sinx=0 in 0 le x le 2pi is

The number of solutions of the equation x^(3)+x^(2)+4x+2sinx=0 in 0 le x le 2pi is

The number of solutions of the equation x^(3)+x^(2)+4x+2sinx=0 in 0 le x le 2pi is