if the equation `sin (pi x^2) - sin(pi x^2 + 2 pi x) = 0` is solved for positive roots, then in the increasing sequence of positive root

The equation |"sin" x| = "sin" x + 3"has in" [0, 2pi]

Solve the equation,sin x=[x],x in[0,2 pi]

Solve the equation sqrt3 cos x + sin x =1 for - 2pi lt x le 2pi.

Solve the equation (sin x + cos x)^(1 + sin2x) = 2 , when -pi le x le pi .

Solve the equation ( sin x + cos x )^(1+ sin 2x )=2 , when -pi le x le pi .

On the domain [-pi,pi] the equation 4 sin^3 x+2 sin^2 x-2 sinx-1=0 possess (A) only one real root(C) four real roots= 0 possess(B) three real roots(D) six real roots

Sum of roots of the equation x ^(4) - 2 x ^(2) sin ^(2)""(pi x)/(2) + 1 = 0 is

The least positive integral solution of "sin" pi (x^(2) + x)-"sin" pi x^(2) = 0 , is

The number of distinct real roots of the equation, |(cos x, sin x , sin x ),(sin x , cos x , sin x),(sinx , sin x , cos x )|=0 in the interval [-pi/4,pi/4] is :