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Find the number of solutions of the equation $2{({\sin ^{ - 1}}x)^2} - {\sin ^{ - 1}}x - 6 = 0$.A) $2$B) $1$C) $0$D) $3$

Last updated date: 16th Jun 2024
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Hint: The given equation is a second-degree equation in ${\sin ^{ - 1}}x$. To find what all values $x$ can take (and thus the number of solutions), first we need to check what all values ${\sin ^{ - 1}}x$ can take. This can be done by solving the quadratic equation by taking ${\sin ^{ - 1}}x = y$(say). Then we can eliminate any values which does not belong to the range of ${\sin ^{ - 1}}x$.

Formula used:
A second-degree equation of the form $a{x^2} + bx + c = 0$ can be solved by
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

The given quadratic equation is $2{({\sin ^{ - 1}}x)^2} - {\sin ^{ - 1}}x - 6 = 0$
We are asked to find the number of solutions.
Here the variable present is $x$. So, we need to find the number of values $x$ can take.
Let, ${\sin ^{ - 1}}x = y$
Then the equation becomes,
$\Rightarrow 2{y^2} - y - 6 = 0$, which is a quadratic equation in $y$.
A second-degree equation of the form $a{x^2} + bx + c = 0$ can be solved by
$\Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Here the variable is $y$.
Also $a = 2,b = - 1,c = - 6$
$\Rightarrow y = \dfrac{{ - ( - 1) \pm \sqrt {{{( - 1)}^2} - 4 \times 2 \times - 6} }}{{2 \times 2}}$
Simplifying we get,
$\Rightarrow y = \dfrac{{1 \pm \sqrt {1 + 48} }}{4} = \dfrac{{1 \pm \sqrt {49} }}{4}$
$\Rightarrow y = \dfrac{{1 \pm 7}}{4}$
This gives $y = \dfrac{8}{4} = 2$ or $y = \dfrac{{ - 6}}{4} = \dfrac{{ - 3}}{2}$
Substituting for $y$ we get,
$\Rightarrow {\sin ^{ - 1}}x = 2{\text{ (or) }}{\sin ^{ - 1}}x = \dfrac{{ - 3}}{2}$
But ${\sin ^{ - 1}}x \in ( - \dfrac{\pi }{2},\dfrac{\pi }{2})$, which is the open interval with limits $- \dfrac{\pi }{2},\dfrac{\pi }{2}$.
Here $\pi$ is in radians, which can be approximated to $3.14$.
This gives
$\Rightarrow \dfrac{\pi }{2} = \dfrac{{3.14}}{2} = 1.57$
So ${\sin ^{ - 1}}x$ cannot take the value $2$, since the value must be less than $1.57$.
Therefore ${\sin ^{ - 1}}x = \dfrac{{ - 3}}{2} \Rightarrow x = \sin (\dfrac{{ - 3}}{2})$, which is the only solution.
This gives the number of solutions is $1$.

$\therefore$ The correct answer is option B.

Note:
The alternative method to solve the quadratic equation without this formula.
$2{y^2} - y - 6 = 0$
By simple rearrangement we get,
$\Rightarrow 2{y^2} - 4y + 3y - 6 = 0$
Taking common factors from first two terms and last two terms,
$\Rightarrow 2y(y - 2) + 3(y - 2) = 0$
$\Rightarrow (y - 2)(2y + 3) = 0$
The product of two terms is zero implies either one is zero.
$\Rightarrow (y - 2) = 0{\text{ (or) }}(2y + 3) = 0$
$\Rightarrow y = 2\left( {{\text{or}}} \right) = \dfrac{{ - 3}}{2}$