<s>[INST] Find minimum value of $8 \cos x + 4 \sin x $ and corresponding value of $x$ Find minimum value of $8 \cos x + 4 \sin x $ and corresponding value of $x$
I used the R method to simplify it - 
$ \sqrt{80} \cos (x-26.565) $ 
Minimum value of that = 
$ \sqrt{80} \cos (x-26.565) = - \sqrt{80}$
$ \cos (x-26.565) = -1$ 
This cosine value lies in the 2nd and 3rd quadrant 
letting $x-26.565 = y$
y reference angle = $ \cos^-1 (-1) = 180$ 
2nd quadrant - $180 - y (ref) = 0 $ 
3rd quadrant - $180 + y(ref) = 360$ 
Therefore , $x = 26.565, 386.565$ 
Why am I wrong ? The minimum value is $206.6$ 
 [/INST] Since
$$
8^2+4^2=80
$$
you know that $8\cos x+4\sin x=\sqrt{80}\cos(x-\alpha)$, for some angle that can be determined by setting $x=0$ and $x=\pi/2$: 
\begin{align}
8&=\sqrt{80}\cos\alpha\\
4&=\sqrt{80}\sin\alpha
\end{align}
Thus the angle $\alpha$ is in the first quadrant and so
$$
\alpha=\arcsin\frac{4}{\sqrt{80}}=\arcsin\frac{1}{\sqrt{5}}
$$
In degrees this is $26.565$. In radians it is $0.464$ (rounding to three decimal digits).
The point where the minimum value $-\sqrt{80}$ is reached is when $x-\alpha$ is the straight angle.  In degrees the value of $x$ is $180+26.565=206.565$.
In radians it is $\pi+0.464=3.605$.
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