Question

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

• A. \N
• B. 1
• C. 2
• D. 4

Answer 1

The correct option is B 1

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