sin3x<sinx
sin3x−sinx<0
Put sin3x=3sinx−4sin3x
3sinx−4sin3x−sinx<0
2sinx−4sin3x<0
2sinx(1−2sin2x)<0
sinxcos2x<0 [cos2x=1−2sin2x]
sinx→+ve cos2x→−ve
xϵ(π4,π2]∪[π2,3π4)=(π4,3π4)
x→−ve cos2x→+ve
xϵ(π,5π4)∪(7π4,2π)
General solution will be xϵ(xπ+π4+nπ+3π4)∪(nπ+π,nπ+5π4)∪(nπ+7π4+nπ+2π)
Where n is even number.