Question

The number of solutions of the equation $si{n}^{-1}6x+si{n}^{-1}6\sqrt{3}x=\frac{-\pi }{2}$ is :

• A. 1
• B. 2
• C. 4

Answer 1

The correct option is A 1

For the equation $si{n}^{-1}6x+si{n}^{-1}6\sqrt{3}x=\frac{-\pi }{2}$, we must have $x<0$ and $-1\le 6x<0$ and $-1<6\sqrt{3}x<0$.
Since for $x>0$, both angles in L.H.S. are in $Q_1$ so their sum cannot be equal to $\frac{-\pi }{2}$ .
Now, $si{n}^{-1}6\sqrt{3}x=\frac{-\pi }{2}-si{n}^{-1}6x=-\frac{\pi }{2}-(\frac{\pi }{2}-co{s}^{-1}6x)=-\pi +co{s}^{-1}6x$
$=-\pi +\pi -si{n}^{-1}\sqrt{1-36x^2}$
$\Rightarrow 6\sqrt{3}x=-\sqrt{1-36x^2}$ $\Rightarrow 108x^2=1-36x^2$
$\Rightarrow x^2=\frac{1}{144}$ $\Rightarrow x=-\frac{1}{12}$ as $x<0$