The inequality $\sin(\frac{\pi}{3}+x)+\sin(\alpha+x)\geq0$ is true for all real numbers $x$

If the value of $\alpha\in[0,2\pi]$ such that the inequality $\sin(\frac{\pi}{3}+x)+\sin(\alpha+x)\geq0$ is true for all real numbers $x$,is equal to $\frac{p\pi}{q}$ where $p$ and $q$ are relatively prime positive integers,find $p$ and $q$

$\sin(\frac{\pi}{3}+x)+\sin(\alpha+x)\geq0$
$\sin\frac{\pi}{3}\cos x+\cos\frac{\pi}{3}\sin x+\sin\alpha\cos x+\cos\alpha\sin x\geq0$

$(\sin \alpha+\frac{\sqrt3}{2})\cos x+(\frac{1}{2}+\cos \alpha)\sin x\geq0$

Let $f(x)=(\sin \alpha+\frac{\sqrt3}{2})\cos x+(\frac{1}{2}+\cos \alpha)\sin x\geq0$, then the range of $f(x)$ is $[-A,A]$, where $A = \sqrt{(\sin \alpha+\frac{\sqrt3}{2})^2+(\frac{1}{2}+\cos \alpha)^2}$. Thus, to ensure that the inequality always hold, we need $\sin \alpha+\frac{\sqrt3}{2}=0$ and $\frac{1}{2}+\cos \alpha=0$. Finally, $\alpha = \frac{4}{3}\pi$
Let me try. You have: $$0 \le \sin(\frac{\pi}{3} + x) + \sin(\alpha + x) = 2\sin (\frac{\pi}{6} + x + \frac{\alpha}{2})\cos (\frac{\pi}{6} - \frac{\alpha}{2})$$
So, you get $\sin (\frac{\pi}{6} + x + \frac{\alpha}{2})$, $\cos (\frac{\pi}{6} - \frac{\alpha}{2})$ is the same sign. From this, you can get the answer.
Here is an alternate solution: you need the inequality to hold for $x=\frac{7\pi}{6}$, so $\sin(\frac{3\pi}{2})+\sin(\alpha+\frac{7\pi}{6})\geq0,\sin(\alpha+\frac{7\pi}{6})\geq 1$ so $\sin(\alpha+\frac{7\pi}{6})=1$, hence $\alpha+\frac{7\pi}{6}=\frac{\pi}{2}+2k\pi$ for some integer $k$, and it's easy enough to find $k$.
• +1 good method,how did you think $x=\frac{7\pi}{6}$?@Wojowu – Vinod Kumar Punia Dec 24 '15 at 10:46
• @VinodKumarPunia I wanted to get $\frac{\pi}{3}+x=\frac{3\pi}{2}$, because that gives $\sin(\frac{\pi}{3}+x)=-1$ and forces $\sin(\alpha+x)=1$. – Wojowu Dec 24 '15 at 10:47