Question

Number of integral solution of the equation $co{s}^{-1}x+co{s}^{-1}(\frac{x}{2}+\frac{1}{2}\sqrt{3-3x^2})=\frac{\pi }{3}$ is -

Answer 1

Let $x=cos\theta$
$co{s}^{-1}\left(cos\theta \right)+co{s}^{-1}(\frac{cos\theta }{2}+\frac{\sqrt{3}}{2}\sqrt{1-co{s}^2\theta })=f\left(\theta \right)$
(i) $x\in [\frac{1}{2},1]\Rightarrow \theta \in [0,\frac{\pi }{3}]$
$f\left(\theta \right)=\theta +co{s}^{-1}\left(cos(\frac{\pi }{3}-\theta )\right)=\frac{\pi }{3}$
(ii) $x\in [0,\frac{1}{2}]\Rightarrow f\left(\theta \right)=2\theta -\frac{\pi }{3}\ne \frac{\pi }{3}$
(iii) $x=0\Rightarrow f\left(\theta \right)=\frac{\pi }{2}+\frac{\pi }{6}\ne \frac{\pi }{3}$
So $x\in [\frac{1}{2},1]\to$ only one integral value.
It should be noted that x can't be negative because range of ${\cos }^{-1}x$ will be from $(0,\pi )$, and for x being negative range will be $(\frac{\pi }{2},\pi )$, so in the above equation we will never get value as $\frac{\pi }{3}$ for the range $(\frac{\pi }{2},\pi )$
So there is only one integral solution.