How to express $-\sqrt{3} \sin x - \cos x$ in the form of $R \cos(x+\theta)$, where $R>0$ and $\theta \in [0^{\circ}, 360^{\circ} ]$ ?
Hence, how to find the maximum value of $y=1+2\sqrt{3} \sin x + 2 \cos x$ and the corresponding value of $x$ for $x \in [0^{\circ}, 360^{\circ} ]$ ?
Any hints?
Your goal in such problems is to express the linear combination $$a\cos\phi +b\sin\phi,$$ with $a,b,\phi\in \mathrm R$ in the form $$R(\sin\psi\cos\phi+\cos\psi\sin\phi),$$ which can be written as $$R\sin(\psi+\phi).$$ You may also use the form of a cosine, whichever is more convenient. If you want a cosine, just note that $\sin (\psi+\phi)=\cos(π/2-\psi-\phi).$
In any case, let's continue. Can we always do this? Yes. First, because we know that $|\sin\psi|,\,|\cos\psi|\le 1,$ we can always achieve this by factoring out $|(a,b)|=\sqrt{a^2+b^2},$ since for all real $a,b$ we have that $a,\,b\le \sqrt{a^2+b^2}.$ If we write $R=\sqrt{a^2+b^2},$ then our linear combination becomes $$R\left(\frac aR\cos\phi+\frac bR\sin\phi\right),$$ so that we may take $\sin\psi =\frac aR$ and $\cos\psi=\frac bR.$ Now we may write $$a\cos\phi +b\sin\phi=R\sin(\psi+\phi),$$ where we know how to calculate $R$ and $\psi.$
To apply this to your specific problem, note that $a=-1$ and $b=-\sqrt 3.$ You should be able to continue to find $R$ and $\psi$ now. Then you can write your expression as $$R\sin(\psi+\phi)=R\cos(π/2-\psi-\phi).$$
For the second part, note that $|a+b+c|\le |a|+|b|+|c|,$ and combine this with the facts that $|\sin y|,\,|\cos y|\le 1$ to bound the second function above. Then determine whether it achieves that bound in the interval $[0,2π].$
Hint: \begin{align}-\sqrt3\sin(x)-\cos(x)&=2\left(-\frac{\sqrt3}2\sin(x)-\frac12\cos(x)\right)\\&=2\bigl(\cos(210^\circ)\sin(x)+\sin(210^\circ)\cos(x)\bigr)\end{align}
$R\cos(x+\theta) = -\sqrt3 \sin(x) - \cos(x)$
Match pattern: $R\cos(x+\theta) = R\cos(x)\cos(\theta) - R\sin(x)\sin(\theta)$
$$R\cos(\theta)=-1$$ $$R\sin(\theta)=\sqrt3$$
$\theta = \tan^{-1}(-\sqrt3) = 120°$
$R={-1\over\cos(\theta)} = 2$
Actually, using arctan to recover angle might not be correct.
It is possible the angle is off 180°. Since R is positive, we are OK.
$y = 1 + 2\sqrt3\sin(x)+2\cos(x) = 1 - 4\cos(x+120°)$
$max(y) = 1 + 4 = 5 → x = 180° - 120° = 60°$
you have $$\sqrt{3}*sin(x)=cos(x)$$ to solve you have $$tan(x)=\frac{1}{\sqrt{3}}$$ but also $$a*sin(x)+b*cos(x)=\sqrt{a^2+b^2}*(\frac{a}{\sqrt{a^2+b^2}}sin(x) +\frac{b}{\sqrt{a^2+b^2}}*cos(x))$$ then $$ \frac{a}{\sqrt{a^2+b^2}}=sin(\phi) and \frac{b}{\sqrt{a^2+b^2}}*cos(x))=cos(\phi)$$
We are looking for $R>0$ and $ \theta$ such that
$$-\sqrt{3} \sin x−\cos x =R \cos(x+θ)$$
for all $x$. For $x=0$ we get
$ \quad -1=R \cos \theta$,
and with $x= \pi/2 = (90^°)$ we derive
$ \quad \sqrt{3}= R \sin \theta.$
Squaring gives $1=R^2 \cos^2 \theta$ and $3=R^2 \sin^2 \theta.$ If we add these equations we derive $R=2.$
Now it is your task to determine $ \theta.$
