Solutions of `sin^(-1)(sinx)=sinx" are if " x in (0,2pi)`

Similar Questions

Explore conceptually related problems

f(x) = sinx - sin2x in [0,pi]

Total number of solutions of sin^(4)x+cos^(4)x=sinx*cosx in [0,2pi] is equal to

Solve the equation , 5sinx+5/(2sinx)-5=2sin^2x+1/2sin^2x if x in (0,pi).

Differentiate sin^(-1)(sinx),\ \ x in [0,\ 2pi]

The number of solutions of sin^(4)x-cos^(2)xsinx+2sin^(2)x+sinx=0 in 0lexle 3pi is

The number of solutions of the equation sinx. Sin2x. Sin3x = 1 in [0, 2pi]

The number of solution of sinx+sin3x+sin5x=0 in the interval {:[(pi)/(2)","3(pi)/(2)]:} is

The solution of sinx+sin5x=sin3x in (0,(pi)/2)