3sinx +4cosx =10 +4x +x^2. Find the number of solutions of the equation

=>3/5 sinx + 4/5 cosx = 1/5(x²+4x+10)Let 3/5 = cos(a) ; so, 4/5 = sin(a)Hence, sinx. cosa + cosx. sina = 1/5(x²+4x+10)=> sin(x+a) = 1/5(x²+4x+10)Now, 1/5(x²+4x+10)=1/5(x²+4x+4+6)=1/5{(x+2)²+6}=1/5(x+2)² + 6/5So min value is 6/5So we have sin(a+x) =6/5Which is not possible as, max value of sin(a+x) is 1.Hence, No solution.