Question

If ${\sin }^{-1}x+{\sin }^{-1}\left(x\sqrt{3}\right)=-\frac{\pi }{2}$, then $x$ is equal to

• A. $\pm \frac{1}{2}$
• B. $-\frac{1}{2}$
• C. $\frac{1}{2}$
• D. None of these

Answer 1

The correct option is B $-\frac{1}{2}$

We have;

${\sin }^{-1}x+{\sin }^{-1}\left(x\sqrt{3}\right)=\frac{-\pi }{2}$

$⟹{\sin }^{-1}[x\sqrt{1-{3x}^2}+x\sqrt{3}\sqrt{1-x^2}]={\sin }^{-1}(-1)$

$⟹x\sqrt{1-{3x}^2}+x\sqrt{3}\sqrt{1-x^2}=-1$

$⟹x\sqrt{1-{3x}^2}=-1-x\sqrt{3}.\sqrt{1-x^2}$

Squaring on both sides,we get;

$⟹x^2(1-{3x}^2)=1+{3x}^2(1-x^2)+2x\sqrt{3}.\sqrt{1-x^2}$

$⟹x^2-3x^4=1+3x^2-3x^4+2x\sqrt{3}.\sqrt{1-x^2)}$

$⟹-(2x^2+1)=2x\sqrt{3}\sqrt{1-x^2}$

Squaring on both sides,we get;

$(2x^2+1{)}^2=4x^2\times 3(1-x^2)$

$⟹4x^2+1+4x^2=12x^2-12x^4$

$⟹16x^4-8x^2+1=0$

$⟹(4x^2-1{)}^2=0$

$⟹4x^2-1=0$

$⟹x^2=\frac{1}{4}⟹x=\pm \frac{1}{2}$

Hence;

$x=-\frac{1}{2}$