If x2+52=x−2cos⁡(m+nx) has at least one real root, then

Given equation isx2+5−2x=−4cos⁡(m+nx)⇒ (x−1)2+4=−4cos⁡(m+nx)LHS≥4 and RHS≤4So, solution is possible only if LHS = RHS = 4∴     x=1 and cos⁡(m+n)=−1.∴     m+n=(2n+1)π,n∈I