Question

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

Answer 1

7

Answer 2

Using the sum-to-product formula $\sin A + \sin B = 2 \sin\left(\frac{A+B}{2}\right) \cos\left(\frac{A-B}{2}\right)$, the inequality becomes $2 \sin\left(\frac{\frac{\pi}{3} + \alpha}{2} + x\right) \cos\left(\frac{\frac{\pi}{3} - \alpha}{2}\right) \ge 0$. For this to hold for all $x$, we must have $\cos\left(\frac{\frac{\pi}{3} - \alpha}{2}\right) = 0$. This implies $\frac{\frac{\pi}{3} - \alpha}{2} = (2n+1)\frac{\pi}{2}$, so $\frac{\pi}{3} - \alpha = (2n+1)\pi$, which gives $\alpha = \frac{\pi}{3} - (2n+1)\pi$. For $\alpha \in [0, 2\pi]$, the only solution is $\alpha = \frac{4\pi}{3}$ (when $n=-1$). Comparing $\frac{4\pi}{3}$ with $\frac{p\pi}{q}$, we get $p=4$ and $q=3$. Thus, $p+q = 4+3=7$.