How do I find m which is a variable of an equation equal to the sum of the first and second derivative?

Question

I saw this problem in a math worksheet. The problem is given below:

Given the function , $$f(x)=\sin(x)+\sqrt3 \cos(x)$$

Find the real number $m$ such that $$f'(x)+f''(x)=\ln(m^2-4m+5)$$

I attempted by simply calculating the first and second derivatives and then tried to to use exponential but I seem to not have any idea about this problem.

Answer 1

$f(x)=Sin (x) + \sqrt 3Cos (x)=2 Sin(x+\pi/3)$

$f'(x)=2 Cos(x+\pi/3)$

$f''(x)=-2Sin(x+\pi/3)$

$m^2-4m+5 =e^{2[Cos(x+\pi/3)-Sin(x+\pi/3)]}=t$

$m=\frac{4±\sqrt {16-4(5-t)}}{2}$

We must have $16 ≥ 4(5-t)$ for real solutions. For example let $16-4(5-t)=0 ⇒ t=1$⇒ $m=2$

Also $e^{2[Cos(x+\pi/3)-Sin(x+\pi/3)]}=1 ⇒{2[Cos(x+\pi/3)-Sin(x+\pi/3)]}=0$

⇒ $Tan (x+\pi/3)=1=Tan(\pi/4)$ ⇒ $x=-\frac{\pi}{12}$

So one solution is $x=-\frac{\pi}{12} $ and $m=2$