The number of solutions of the equation x3+2x2+5x+2cos⁡x=0 in [0,2π]is

Let f(x)=x3+2x2+5x+2cos ⁡x∴ f′(x)=3x2+4x+5−2sin xNow, the least value of 3x2+4x+5 is −D4a=−(4)2−4(3)(5)4(3)=113and the greatest value of 2 sin x is 2. Therefore 3x2+4x+5>2sin ⁡x∀x∈Ror f′(x)=3x2+4x+5−2sin ⁡x>0∀x∈RThus, f(x) is strictly an increasing function. Also, f(0)=2 and f(2π)>0,Thus, for the given interval , f (x) never becomes zero. Hence, the number of roots is zero.