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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 $206.6$

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Your method is perfectly fine, you just made a mistake at the end.

$\cos(u)=-1\iff u\equiv 180°\pmod{360°}$

Here $u=x-x_0$ so you should get $x\equiv 180°+x_0\pmod{360°}$

Applying to $x_0=26.565°$ you get $x = 206.565°$

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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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