How to find max integer value of $6\sin(x)-8\cos(x)$ without using derivative I have $6\sin(x)-8\cos(x)$ and want to find its maximum value. If we use derivative and assuming that it is equal to 0, we get $6\cos(x)+8\sin(x)=0$ which implies that $\tan(x)=-\cfrac{3}{4}$ and $x = 143^\circ$ and we can conclude that the maximum value is equal to $10$ . How can I find the maximum integer value of $6\sin(x)-8\cos(x)$ without using derivative?

Write 
$$f = 6 \sin(x) - 8 \cos(x) = \sqrt{6^2 + 8^2} (\frac {6}{\sqrt{6^2 + 8^2}} \sin(x) + \frac {8}{\sqrt{6^2 + 8^2}} \cos(x) ) $$
and now let
$$\frac {6}{\sqrt{6^2 + 8^2}} = \sin(\alpha)$$
then (remember $\sin(z)^2 + \cos(z)^2 = 1$ for any $z$)
$$\frac {8}{\sqrt{6^2 + 8^2}} = \cos(\alpha)$$
And we can write
$$f = 10 \;(\;\sin(\alpha)\sin(x) - \cos(\alpha) \cos(x)\; ) = - 10 \cos(x+\alpha) $$
where we have used the trigonometric addition theorem.     
Hence the maximum is obviously $10$.
EDIT
As a bonus let us determine the values of $x$ where $f$ has its maximum (and its minimum).
From the previous result we have to look for the extrema of $\cos(z)$ which leads us to
$$f\to -10:  x =2 n \pi - \alpha $$ 
$$f\to +10:  x=(2 n+1) \pi - \alpha$$
Where
$$\alpha = \arctan(\frac{3}{4})\;\simeq 0.6435010.643501 $$
and $n$ integer.
The extrema closest to the origin are at
$$x_{min} = - \alpha\;\simeq -0.6435010.643501 $$
$$x_{max} = \pi - \alpha \; \simeq 2.498091544796$$
