Question

The number of real solutions of the equation $ta{n}^{-1}\sqrt{x(x+1)}+si{n}^{-1}\sqrt{x^2+x+1}=\frac{\pi }{2}$ is

• A. zero
• B. one
• C. two
• D. infinite

Answer 1

The correct option is C

two

Clearly, $x(x+1)\ge 0$ and $x^2+x+1\le 1$. Together they inmly x(x+1)=0
$\therefore$ x = 0, -1
When x = 0,
$LHS=ta{n}^{-1}0+si{n}^{-1}=\frac{\pi }{2}$
When x=-1,
$LHS=ta{n}^{-1}0+si{n}^{-1}\sqrt{1-1+1}=0+si{n}^{-1}1=\frac{\pi }{2}$,
Thus, 2 solution