MCQ
Solve $\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2},$ then $x$ is equal to
- A$\frac{1}{2}$
- B$0, \frac{1}{2}$
- C$1, \frac{1}{2}$
- ✓$0$
$\Rightarrow-2 \sin ^{-1} x=\frac{\pi}{2}-\sin ^{-1}(1-x)$
$\Rightarrow-2 \sin ^{-1} x=\cos ^{-1}(1-x)$ $\dots(1)$
Let $\sin ^{-1} x=\theta \Rightarrow \sin \theta=x \Rightarrow \cos \theta=\sqrt{1-x^{2}}$
$\therefore \theta=\cos ^{-1}(\sqrt{1-x^{2}})$
$\therefore \sin ^{-1} x=\cos ^{-1}(\sqrt{1-x^{2}})$
Therefore, from equation $( 1 )$, we have
$-2 \cos ^{-1}(\sqrt{1-x^{2}})=\cos ^{-1}(1-x)$
Let, $x=\sin y .$ Then, we have:
$-2 \cos ^{-1}(\sqrt{1-\sin ^{2} y})=\cos ^{-1}(1-\sin y)$
$\Rightarrow-2 \cos ^{-1}(\cos y)=\cos ^{-1}(1-\sin y)$
$\Rightarrow-2 y=\cos ^{-1}(1-\sin y)$
$\Rightarrow 1-\sin y=\cos (-2 y)=\cos 2 y$
$\Rightarrow 1-\sin y=1-2 \sin ^{2} y$
$\Rightarrow 2 \sin ^{2} y-\sin y=0$
$\Rightarrow \sin y(2 \sin y-1)=0$
$\Rightarrow \sin y=0$ or $\frac{1}{2}$
$\therefore x=0$ OR $x=\frac{1}{2}$
When $x=\frac{1}{2},$ it can be observed that
L.H.S. $=\sin ^{-1}\left(1-\frac{1}{2}\right)-\sin ^{-1} \frac{1}{2}$
$=\sin ^{-1}\left(\frac{1}{2}\right)-2 \sin ^{-1} \frac{1}{2}$
$=-\sin ^{-1} \frac{1}{2}$
$=\frac{\pi}{6} \neq \frac{\pi}{2} \neq R . H . S$
$\therefore x=\frac{1}{2}$ is not a solution of the given equation. Thus, $x=0$
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where $A = {\sin ^2}\alpha - \sin \alpha + \frac{1}{4}$
and $B = {\tan ^2}\alpha + \frac{2}{{\sqrt 3 }}\tan \alpha + \frac{1}{3}$ , then the number of value $(s)$ of $\alpha $ in $\left[ { - \frac{{3\pi }}{2},2\pi } \right]$ is - (where $sgnx$ denotes signum function of $x$ )