`sqrt([sin2x]-[cos2x])` Find the range

f(x)=sqrt([sin2x]-[cos2x]) :[.]GIF, then range of f(x)

Let f(x)=sqrt([sin2x]-[cos2x]) (where II denotes the greatest integer function then the range of f(x) will be

if: f(x)=(sin x)/(sqrt(1+tan^(2)x))-(cos x)/(sqrt(1+cot^(2)x)), then find the range of f(x)

sin^(-1)sqrt(x^(2)+x+1) Find the range

For x in(-pi,pi) find the value of x for which the given equation (sqrt(3)sin x+cos x)^(sqrt(sqrt(3)sin2x-cos2x+2))=4 is satisfied.

For which values of a does the equation 4sin(x+pi/3)cos(x-pi/6)=a^2+sqrt(3)sin2x-cos2x have solution? Find the solution for a=0.

For which values of a does the equation 4sin(x+pi/3)cos(x-pi/6)=a^2+sqrt(3)sin2x-cos2x have solution? Find the solution for a-0.

If f(x)=|(10(sqrt(cos x)-sqrt(sin x))(sqrt(cos x)+ sqrt(sin x)))/(cos x+ sin x)| then number of natural numbers in the range of f(x) is

If int (sin2x-cos2x)dx=1/sqrt2sin(2x-a)+b then