The maximum of 3 sin x + 4 cos x

Critical points. $f'(x)=0\iff\tan x=\frac34$. The solutions in $[0,2\pi)$ are $x_0=\arctan\frac34$ (first quadrant) and $x_0+\pi$ (third quadrant).

Values. At $x_0$, $\sin x_0=\frac35$, $\cos x_0=\frac45$ (a $3$-$4$-$5$ triangle), so $f(x_0)=\frac95+\frac{16}5=5$. At $x_0+\pi$ both $\sin$ and $\cos$ change sign: $f=-5$.

Conclusion. $f$ is continuous and $2\pi$-periodic, so it attains its extreme values on $[0,2\pi]$, and since it is differentiable everywhere these are attained at critical points: maximum $5$, minimum $-5$.

Phase shift. $3\sin x+4\cos x=5\sin(x+\varphi)$ with $\cos\varphi=\frac35$, $\sin\varphi=\frac45$: values in $[-5,5]$, in agreement.

Remark. In general, $a\sin x+b\cos x$ has maximum $\sqrt{a^2+b^2}$. This is also the Cauchy–Schwarz inequality $a\cos t+b\sin t\le\sqrt{a^2+b^2}$.