Question

Find the real solution of

$ta{n}^{-1}\sqrt{x(x+1)}+si{n}^{-1}\sqrt{x^2+x+1}=\frac{x}{2}$.

Answer 1

We have, $ta{n}^{-1}\sqrt{x(x+1)}+si{n}^{-1}\sqrt{x^2+x+1}=\frac{x}{2}\cdots \left(i\right)$

Let $si{n}^{-1}\sqrt{x^{+}x+1}=\theta \sqrt{-x^2+x+1}$

$\Rightarrow sin\theta =\sqrt{\frac{x^2+x+1}{1}}\sqrt{-x^2-x}$

$\Rightarrow tan\theta =\frac{\sqrt{x^2+x+1}}{\sqrt{-x^2-x}}[\because tan\theta =\frac{sin\theta }{cos\theta }]\because \theta =ta{n}^{-1}\frac{\sqrt{x^2+x+1}}{\sqrt{-x^2-x}}=si{n}^{-1}\sqrt{x^2+x+1}$

On putting the value of $\theta$ in Eq. (i), we get

$ta{n}^{-1}\sqrt{x(x+1)}+ta{n}^{-1}\frac{\sqrt{x^2+x+1}}{\sqrt{-x^2-x}}=\frac{x}{2}$

We know that, $ta{n}^{-1}x+ta{n}^{-1}y=ta{n}^{-1}\left(\frac{x+y}{1-xy}\right)xy<1$

$\because ta{n}^{-1}\left[\frac{\sqrt{x(x+1)}+\sqrt{\frac{x^2+x+1}{-x^2-x}}}{1-\sqrt{x(x+1)}.\frac{\sqrt{x^2+x+1}}{-x^2-x}}\right]=\frac{\pi }{2}\Rightarrow ta{n}^{-1}\left[\frac{\sqrt{x^2+x}+\sqrt{\frac{x^2+x+1}{-1(x^2+x)}}}{1-\sqrt{(x^2+x)}.\frac{(x^2+x+1)}{-1(x^2+x)}}\right]=\frac{\pi }{2}\Rightarrow \frac{x^2+x+\sqrt{-1(x^2+x+1)}}{[1-\sqrt{-(x^2+x+1)}]\sqrt{(x^2+x)}}=tan\frac{\pi }{2}=\frac{1}{0}$

$\Rightarrow [1-\sqrt{-(x^2+x+1)}]\sqrt{(x^2+x)}=0\Rightarrow -(x^2+x+1)=1or{x}^2+x=0\Rightarrow -x^2-x-1=1orx(x+1)=0\Rightarrow x^2+x+2=0orx(x+1)=0\therefore x=\frac{-1\pm \sqrt{1-4\times 2}}{2}\Rightarrow x=0orx=-1$

For real solution, we have x = 0, -1.