Solving 2\sin(x+30^{\circ})=\cos(x+150^{\circ}) for x between 0^{\circ} \text{ and } 360^{\circ}

Let P(x, y) be the terminal point on the unit circle determined by t. Then $\sin t=,\cos t=,\text{and}\tan t=$.

In any triangle is $\sin A+\sin B+\sin C=\frac{3\sqrt{3}}{2}$ always Well, I came with an interesting proof. But I just want to verify it Applied at function $y=\sin x$ We have, $\sin (\frac{A+B+C}{2})\ge \frac{\sin A+\sin B+\sin C}{3}$ From here we will get $\sin A+\sin B+\sin C\le \frac{3\sqrt{3}}{2}$ Also by A.M. $\ge$ G.M. in an acute angled triangle $\frac{\sin A+\sin B+\sin C}{3}\ge \sqrt[3]{\sin A\sin B\sin C}$ $\Rightarrow \sin A+\sin B+\sin C\ge 3(\sqrt[3]{\sin A\sin B\sin C})$ $\Rightarrow \sin A+\sin B+\sin C\ge 3(\frac{\sqrt{3}}{2})=\frac{3\sqrt{3}}{2}>2$ and from this I get $\sin A+\sin B+\sin C\ge \frac{3\sqrt{3}}{2}$

I have an impulse train given by
${\displaystyle \frac{1}{R+1}+\frac{\sum _{k=1}^R\cos \left(\frac{2k\pi x}{R+1}\right)}{R+1}}$
It seems obvious to me that, for x=0, the function returns 1. This is because ${\displaystyle \cos \left(0\right)=1}$, and we therefore end up with ${\displaystyle \frac{1}{R+1}+\frac{R}{R+1}=\frac{R+1}{R+1}=1}$
However indeterminate result at x=0. Usually this means there is a division by 0 somewhere. But I can't see any reason for this function to produce an indeterminate result.

Proving ${\displaystyle \sin 4\theta =4{\cos }^3\theta \sin \theta -4\cos \theta {\sin }^3\theta }$
Using De Moivre's formula:
${\displaystyle {(\cos \theta +i\sin \theta )}^4=\cos 4\theta +i\sin 4\theta }$

How can one tell the period of ${\displaystyle \sin \left(x^{\frac{3}{2}}\right)}$ ? Is it ${\displaystyle {\left(2\pi \right)}^{\frac{2}{3}}}$?

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${\displaystyle 4{\cos }^2\frac{\pi }{5}-2\cos \frac{\pi }{5}-1=0}$

Tips on evaluating ${\displaystyle \underset{x\to 0}{lim}\frac{\sqrt{1-\cos \left(x^2\right)}}{1-\cos x}}$?