The number of solutions of the equation x3+2x2+5x+2cosx=0 in [0,2π] is
Solution : x3+2x2+5x+2cosx=0 x3+2x2+5x=−2cosx y=x(x2+2x+5)=−2cosx y=x(x2+2x+5) if x∈[0,2π] then number of solution is zero.
The number of solutions of the equation |cosx–sinx|=2cosx, xϵ[0,2π] is